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Special Interals



Why Memorize Special Integrals?
Power Functions
Exponential Functions
Trigonometric Functions
Inverse Trigonometric Forms
Logarithmic Integrals
Integrals Leading to Logarithms
Master Reference: All Special Integrals at a Glance
Common Pitfalls
Special Integrals FAQ



Formulas Worth Knowing


Certain integrals appear so frequently that recognizing them on sight saves considerable effort. These standard forms serve as endpoints—the targets that techniques aim to reach.

Each formula below reverses a known derivative. The integral of cos⁡x\cos x is sin⁡x\sin x because the derivative of sin⁡x\sin x is cos⁡x\cos x. The integral of 1/(1+x2)1/(1 + x^2) is arctan⁡x\arctan x because the derivative of arctan⁡x\arctan x is 1/(1+x2)1/(1 + x^2).

Memorizing these forms accelerates computation and provides reference points for more complex integrals. When a technique transforms an integral into one of these patterns, the work is essentially done.

Key Terms

Antiderivative— each formula reverses a known derivative
Indefinite Integral— the notation ∫f(x) dx=F(x)+C\int f(x)\,dx = F(x) + C
Integrand— recognizing the integrand's form selects the formula

See All Calculus Definitions →


Why Memorize Special Integrals?


    Integration techniques transform unfamiliar integrals into recognizable ones. These standard forms are the targets.

    Knowing them provides:

  • Speed—instant recognition eliminates derivation
  • Building blocks—complex integrals reduce to combinations of simple ones
  • Verification—differentiate to confirm correctness
  • Pattern anchors—guide technique selection by identifying what form to aim for

    • Each formula inverts a differentiation rule. Mastery of derivatives implies knowledge of these integrals.

Power Functions


The power rule for integration:

Power Rule (Integrals)
∫xn dx=xn+1n+1+Cn≠−1\int x^n\, dx = \frac{x^{n+1}}{n+1} + C \quad n \neq -1
Learn more about this formula: Power Rule (Integrals) →

The exponent increases by one; divide by the new exponent.

The exception n=−1n = -1 is critical:

Reciprocal Antiderivative
∫1x dx=ln⁡∣x∣+C\int \frac{1}{x}\, dx = \ln|x| + C
Learn more about this formula: Reciprocal Antiderivative →


The absolute value ensures validity for negative xx. For x>0x > 0, the antiderivative is ln⁡x\ln x; for x<0x < 0, it is ln⁡(−x)\ln(-x). Both cases combine into ln⁡∣x∣\ln|x|.

Pitfall: Using ln for every reciprocal
Writing ∫1x2 dx=ln⁡(x2)+C\int \frac{1}{x^2}\, dx = \ln(x^2) + C gives a function whose derivative is 2x\frac{2}{x}, not 1x2\frac{1}{x^2}. As x−2x^{-2}, the power rule gives x−1−1+C=−1x+C\frac{x^{-1}}{-1} + C = -\frac{1}{x} + C.

The logarithm belongs to the exponent −1-1 alone. Other reciprocals of powers, such as 1x2=x−2\frac{1}{x^2} = x^{-2} or 1x=x−1/2\frac{1}{\sqrt{x}} = x^{-1/2}, are first rewritten with a negative exponent and then fall under the power rule, so their antiderivatives are again powers of xx.
1/x and ln|x| on both sides of 0: ln|x| has slope 1/x for negative x tooxy−4−3−2−11234−3−2−1123y = 1/xln(−x)ln xslope of ln|x| at −2: −1/2ln|x| is an antiderivative of 1/x on both sides of 0
Why the absolute value in ln|x|

1/x is defined on both sides of 0, but ln x only for x > 0. On the left the antiderivative is ln(−x), the mirror image of ln x, and the two halves together are ln|x|. At x = −2 the left branch is falling with slope −1/2, which is exactly 1/x there. Writing only ln x would leave the negative half of 1/x without an antiderivative.

The two halves can even carry different constants: 0 is not in the domain, so nothing links them.

Exponential Functions


The exponential function is its own antiderivative:

Exponential Antiderivative
∫ex dx=ex+C\int e^x\, dx = e^x + C
Learn more about this formula: Exponential Antiderivative →

For other bases:

General Exponential Antiderivative
∫ax dx=axln⁡a+Ca>0, a≠1\int a^x\, dx = \frac{a^x}{\ln a} + C \quad a > 0, \, a \neq 1
Learn more about this formula: General Exponential Antiderivative →


The factor 1/ln⁡a1/\ln a compensates for the chain rule in (ax)′=axln⁡a(a^x)' = a^x \ln a.

With a linear argument:

Exponential Functions
∫ekx dx=ekxk+C,k≠0\int e^{kx}\, dx = \frac{e^{kx}}{k} + C, \quad k \neq 0

Pitfall: Forgetting the 1/k
Writing ∫e3x dx=e3x+C\int e^{3x}\, dx = e^{3x} + C gives a function whose derivative is 3e3x3e^{3x}, three times the integrand. Dividing by the inner constant gives e3x3+C\frac{e^{3x}}{3} + C.

The factor 1k\frac{1}{k} does the same job as 1ln⁡a\frac{1}{\ln a} above. Differentiating ekxe^{kx} by the chain rule brings down the constant kk, the derivative of the inner kxkx, and the antiderivative divides that constant back out. Any linear argument kx+bkx + b behaves the same way, since its derivative is the constant kk.

Trigonometric Functions


Basic trigonometric integrals:

Antiderivative of Sine
∫sin⁡x dx=−cos⁡x+C\int \sin x\, dx = -\cos x + C
Learn more about this formula: Antiderivative of Sine →

Antiderivative of Cosine
∫cos⁡x dx=sin⁡x+C\int \cos x\, dx = \sin x + C
Learn more about this formula: Antiderivative of Cosine →

Antiderivative of Sec Squared
∫sec⁡2x dx=tan⁡x+C\int \sec^2 x\, dx = \tan x + C
Learn more about this formula: Antiderivative of Sec Squared →

Antiderivative of Csc Squared
∫csc⁡2x dx=−cot⁡x+C\int \csc^2 x\, dx = -\cot x + C
Learn more about this formula: Antiderivative of Csc Squared →

Antiderivative of Sec Tan
∫sec⁡xtan⁡x dx=sec⁡x+C\int \sec x \tan x\, dx = \sec x + C
Learn more about this formula: Antiderivative of Sec Tan →

Antiderivative of Csc Cot
∫csc⁡xcot⁡x dx=−csc⁡x+C\int \csc x \cot x\, dx = -\csc x + C
Learn more about this formula: Antiderivative of Csc Cot →


Pitfall: Integrating sin x to cos x
Writing ∫sin⁡x dx=cos⁡x+C\int \sin x\, dx = \cos x + C gives a function whose derivative is −sin⁡x-\sin x, the integrand with the wrong sign. The antiderivative is −cos⁡x+C-\cos x + C.

Each reverses a standard derivative. The negative signs in sin⁡x\sin x and csc⁡2x\csc^2 x integrals reflect the negatives in (cos⁡x)′=−sin⁡x(\cos x)' = -\sin x and (cot⁡x)′=−csc⁡2x(\cot x)' = -\csc^2 x.
Integrand Antiderivative Reverses derivative of
sin x −cos x + C cos x (which gives −sin x)
cos x sin x + C sin x
sec² x tan x + C tan x
csc² x −cot x + C cot x (which gives −csc² x)
sec x · tan x sec x + C sec x
csc x · cot x −csc x + C csc x (which gives −csc x cot x)

Inverse Trigonometric Forms


These integrals produce inverse trigonometric functions:

Arctangent Form
∫11+x2 dx=arctan⁡x+C\int \frac{1}{1 + x^2}\, dx = \arctan x + C
Learn more about this formula: Arctangent Form →

Arcsine Form
∫11−x2 dx=arcsin⁡x+C\int \frac{1}{\sqrt{1 - x^2}}\, dx = \arcsin x + C
Learn more about this formula: Arcsine Form →

Arcsecant Form
∫1xx2−1 dx=arcsec⁡∣x∣+C\int \frac{1}{x\sqrt{x^2 - 1}}\, dx = \operatorname{arcsec}|x| + C
Learn more about this formula: Arcsecant Form →


More generally, with constant a>0a > 0:

Arctangent Form
∫1a2+x2 dx=1aarctan⁡xa+C\int \frac{1}{a^2 + x^2}\, dx = \frac{1}{a}\arctan\frac{x}{a} + C
Learn more about this formula: Arctangent Form →

Arcsine Form
∫1a2−x2 dx=arcsin⁡xa+C\int \frac{1}{\sqrt{a^2 - x^2}}\, dx = \arcsin\frac{x}{a} + C
Learn more about this formula: Arcsine Form →


Pitfall: Dropping the 1/a in the arctangent form
Writing ∫14+x2 dx=arctan⁡x2+C\int \frac{1}{4 + x^2}\, dx = \arctan\frac{x}{2} + C gives a function whose derivative is 24+x2\frac{2}{4 + x^2}, twice the integrand. With a=2a = 2 the answer is 12arctan⁡x2+C\frac{1}{2}\arctan\frac{x}{2} + C.

Both general forms reduce to the base forms through the substitution x=aux = au, dx=a dudx = a\, du. In the arctangent form the denominator becomes a2(1+u2)a^2(1 + u^2), and one factor aa is left over after dxdx is replaced. In the arcsine form the radical becomes a1−u2a\sqrt{1 - u^2}, and the aa cancels completely, which is why that form has no factor in front.
Integrand Antiderivative Note
1 / (1 + x²) arctan x + C base form (a = 1)
1 / √(1 − x²) arcsin x + C base form (a = 1); valid on |x| < 1
1 / (x · √(x² − 1)) arcsec |x| + C absolute value handles both x > 1 and x < −1
1 / (a² + x²) (1/a) arctan(x/a) + C generalization with a > 0
1 / √(a² − x²) arcsin(x/a) + C generalization with a > 0; valid on |x| < a

Logarithmic Integrals


The logarithm itself requires integration by parts:

Antiderivative of Natural Log
∫ln⁡x dx=xln⁡x−x+C\int \ln x\, dx = x\ln x - x + C
Learn more about this formula: Antiderivative of Natural Log →

A crucial pattern recognizes when integrands have the form f′(x)/f(x)f'(x)/f(x):

Logarithmic Derivative Pattern
∫f′(x)f(x) dx=ln⁡∣f(x)∣+C\int \frac{f'(x)}{f(x)}\, dx = \ln|f(x)| + C
Learn more about this formula: Logarithmic Derivative Pattern →


This pattern appears frequently in disguise.

Example
∫tan⁡x dx=∫sin⁡xcos⁡x dx=−ln⁡∣cos⁡x∣+C\int \tan x\, dx = \int \frac{\sin x}{\cos x}\, dx = -\ln|\cos x| + C
Here f(x)=cos⁡xf(x) = \cos x and f′(x)=−sin⁡xf'(x) = -\sin x, so the integrand is −f′(x)/f(x)-f'(x)/f(x).

Pitfall: Using ln without the derivative on top
Writing ∫1x2+1 dx=ln⁡(x2+1)+C\int \frac{1}{x^2 + 1}\, dx = \ln(x^2 + 1) + C gives a function whose derivative is 2xx2+1\frac{2x}{x^2 + 1}. The numerator 11 is not the derivative of x2+1x^2 + 1; the integral is arctan⁡x+C\arctan x + C.

Before the pattern is used, the numerator is compared with the derivative of the denominator. A constant factor between them does no harm, since it can be moved outside the integral: in ∫xx2+1 dx\int \frac{x}{x^2 + 1}\, dx the derivative of the denominator is 2x2x, so the integral is 12ln⁡(x2+1)+C\frac{1}{2}\ln(x^2 + 1) + C. In the example above the factor is −1-1, which gives the minus sign in −ln⁡∣cos⁡x∣-\ln|\cos x|.

Integrals Leading to Logarithms


Several standard integrals produce logarithms:

Antiderivative of Tangent
∫tan⁡x dx=−ln⁡∣cos⁡x∣+C=ln⁡∣sec⁡x∣+C\int \tan x\, dx = -\ln|\cos x| + C = \ln|\sec x| + C
Learn more about this formula: Antiderivative of Tangent →

Antiderivative of Cotangent
∫cot⁡x dx=ln⁡∣sin⁡x∣+C\int \cot x\, dx = \ln|\sin x| + C
Learn more about this formula: Antiderivative of Cotangent →

Antiderivative of Secant
∫sec⁡x dx=ln⁡∣sec⁡x+tan⁡x∣+C\int \sec x\, dx = \ln|\sec x + \tan x| + C
Learn more about this formula: Antiderivative of Secant →

Antiderivative of Cosecant
∫csc⁡x dx=−ln⁡∣csc⁡x+cot⁡x∣+C=ln⁡∣csc⁡x−cot⁡x∣+C\int \csc x\, dx = -\ln|\csc x + \cot x| + C = \ln|\csc x - \cot x| + C
Learn more about this formula: Antiderivative of Cosecant →


The sec⁡x\sec x and csc⁡x\csc x integrals are less obvious—they require multiplying by clever forms of 11 to create the f′/ff'/f pattern.

Master Reference: All Special Integrals at a Glance


The formulas covered in the sections above all share the same role: standard endpoints that techniques aim to reach. The master reference below collects every formula on this page in one place, grouped by family so that scanning by integrand form returns the matching antiderivative directly. Use it as a lookup card while working integrals — the section above each row supplies the derivation and the context.
Family Integrand Antiderivative (+ C)
Power xn  (n ≠ −1) xn+1 / (n + 1)
Power 1 / x ln |x|
Exponential ex ex
Exponential ax  (a > 0, a ≠ 1) ax / ln a
Exponential ekx  (k ≠ 0) ekx / k
Trigonometric sin x −cos x
Trigonometric cos x sin x
Trigonometric sec² x tan x
Trigonometric csc² x −cot x
Trigonometric sec x · tan x sec x
Trigonometric csc x · cot x −csc x
Inverse trig 1 / (1 + x²) arctan x
Inverse trig 1 / √(1 − x²) arcsin x
Inverse trig 1 / (x · √(x² − 1)) arcsec |x|
Logarithmic ln x x ln x − x
Logarithmic pattern f'(x) / f(x) ln |f(x)|
Trig → log tan x −ln |cos x|  =  ln |sec x|
Trig → log cot x ln |sin x|
Trig → log sec x ln |sec x + tan x|
Trig → log csc x −ln |csc x + cot x|  =  ln |csc x − cot x|
DownloadPNG2×PDF
Before you read
Twenty integrals in seven families
The integrals on this page group by the kind of integrand. Each row shows the integrand and its antiderivative; add + C to every one.
  • L0Special integrals
    • L1Power
      • L2xⁿn ≠ −1xⁿ⁺¹ / (n + 1)
      • L21 / xln |x|
    • L1Exponential
      • L2eˣeˣ
      • L2aˣa > 0, a ≠ 1aˣ / ln a
      • L2eᵏˣk ≠ 0eᵏˣ / k
    • L1Trigonometric
      • L2sin x−cos x
      • L2cos xsin x
      • L2sec² xtan x
      • L2csc² x−cot x
      • L2sec x · tan xsec x
      • L2csc x · cot x−csc x
    • L1Inverse trigonometric
      • L21 / (1 + x²)arctan x
      • L21 / √(1 − x²)arcsin x
      • L21 / (x · √(x² − 1))arcsec |x|
    • L1Logarithmic
      • L2ln xx ln x − x
    • L1Logarithmic pattern
      • L2f′(x) / f(x)ln |f(x)|
    • L1Trigonometric to logarithm
      • L2tan xln |sec x|
      • L2cot xln |sin x|
      • L2sec xln |sec x + tan x|
      • L2csc xln |csc x − cot x|
Takeaway
Where the logarithms come from
Logarithms appear whenever the integrand has the form f′(x) / f(x): 1/x, tan x, cot x, sec x and csc x all follow that pattern.
Special integrals by family·/calculus/integrals/specialLearn Math Class

Common Pitfalls


Five mistakes cause many of the wrong answers when these formulas are used. Each is marked where it happens in the sections above; here each is worked through in full, with the wrong and the right working side by side.

Using ln for every reciprocal

The antiderivative ln⁡∣x∣\ln|x| of 1x\frac{1}{x} is carried over to other reciprocals, so 1x2\frac{1}{x^2} or 1x3\frac{1}{x^3} is integrated to a logarithm as well.

Wrong working: ∫1x2 dx=ln⁡(x2)+C\int \frac{1}{x^2}\, dx = \ln(x^2) + C.

Right working: 1x2=x−2\frac{1}{x^2} = x^{-2}, so ∫x−2 dx=x−1−1+C=−1x+C\int x^{-2}\, dx = \frac{x^{-1}}{-1} + C = -\frac{1}{x} + C.

Why it happens: 1x\frac{1}{x} is the reciprocal met most often, and its integral is the one that gets memorised. Written as a fraction, 1x2\frac{1}{x^2} hides its exponent, so it is not obvious that only the exponent −1-1 is the exception.

How to check: on [1,2][1, 2] the integrand 1x2\frac{1}{x^2} lies between 14\frac{1}{4} and 11, so ∫121x2 dx\int_1^2 \frac{1}{x^2}\, dx lies between 14\frac{1}{4} and 11. The power rule gives [−1x]12=−12+1=12\left[-\frac{1}{x}\right]_1^2 = -\frac{1}{2} + 1 = \frac{1}{2}; the logarithm gives ln⁡4−ln⁡1≈1.386\ln 4 - \ln 1 \approx 1.386, which is too large.

Forgetting the 1/k

The exponential ekxe^{kx} is treated like exe^x, and its antiderivative is written down as the function itself, without dividing by kk.

Wrong working: ∫e3x dx=e3x+C\int e^{3x}\, dx = e^{3x} + C.

Right working: ∫e3x dx=e3x3+C\int e^{3x}\, dx = \frac{e^{3x}}{3} + C, because ddx[e3x3]=3e3x3=e3x\frac{d}{dx}\left[\frac{e^{3x}}{3}\right] = \frac{3e^{3x}}{3} = e^{3x}.

Why it happens: exe^x is its own antiderivative, and the rule "the exponential stays the same" is applied to every exponential. The factor 33 from the chain rule only shows up when the answer is differentiated.

How to check: differentiate the answer. ddx[e3x]=3e3x\frac{d}{dx}[e^{3x}] = 3e^{3x}, three times the integrand, so the answer must be divided by 33.

Integrating sin x to cos x

Because the derivative of sin⁡x\sin x is cos⁡x\cos x with no sign, the integral of sin⁡x\sin x is taken to be cos⁡x\cos x with no sign as well.

Wrong working: ∫sin⁡x dx=cos⁡x+C\int \sin x\, dx = \cos x + C.

Right working: (cos⁡x)′=−sin⁡x(\cos x)' = -\sin x, so (−cos⁡x)′=sin⁡x(-\cos x)' = \sin x, and ∫sin⁡x dx=−cos⁡x+C\int \sin x\, dx = -\cos x + C.

Why it happens: between sin⁡x\sin x and cos⁡x\cos x, the minus sign appears in only one direction of differentiation, and it is easy to attach it to the wrong one.

How to check: on [0,π][0, \pi] the integrand sin⁡x\sin x is never negative, so ∫0πsin⁡x dx\int_0^\pi \sin x\, dx must be positive. The right answer gives −cos⁡π+cos⁡0=1+1=2-\cos \pi + \cos 0 = 1 + 1 = 2; the wrong one gives cos⁡π−cos⁡0=−2\cos \pi - \cos 0 = -2.

Dropping the 1/a in the arctangent form

In ∫1a2+x2 dx\int \frac{1}{a^2 + x^2}\, dx, the argument of the arctangent is correctly changed to xa\frac{x}{a}, but the factor 1a\frac{1}{a} in front is left off.

Wrong working: ∫14+x2 dx=arctan⁡x2+C\int \frac{1}{4 + x^2}\, dx = \arctan\frac{x}{2} + C.

Right working: with x=2ux = 2u and dx=2 dudx = 2\, du, ∫2 du4(1+u2)=12arctan⁡u+C=12arctan⁡x2+C\int \frac{2\, du}{4(1 + u^2)} = \frac{1}{2}\arctan u + C = \frac{1}{2}\arctan\frac{x}{2} + C.

Why it happens: the two general forms are learned together, and the arcsine form arcsin⁡xa\arcsin\frac{x}{a} has no factor in front. The arctangent form is then remembered in the same shape.

How to check: on [0,2][0, 2] the integrand 14+x2\frac{1}{4 + x^2} lies between 18\frac{1}{8} and 14\frac{1}{4}, so ∫0214+x2 dx\int_0^2 \frac{1}{4 + x^2}\, dx lies between 14\frac{1}{4} and 12\frac{1}{2}. The right answer gives 12arctan⁡1=π8≈0.393\frac{1}{2}\arctan 1 = \frac{\pi}{8} \approx 0.393; the wrong one gives arctan⁡1=π4≈0.785\arctan 1 = \frac{\pi}{4} \approx 0.785, outside that range.

Using ln without the derivative on top

Any fraction with a function in the denominator is integrated to the logarithm of that denominator, whether or not the numerator is its derivative.

Wrong working: ∫1x2+1 dx=ln⁡(x2+1)+C\int \frac{1}{x^2 + 1}\, dx = \ln(x^2 + 1) + C.

Right working: 11+x2\frac{1}{1 + x^2} is the arctangent form, so ∫1x2+1 dx=arctan⁡x+C\int \frac{1}{x^2 + 1}\, dx = \arctan x + C. The logarithm belongs to ∫xx2+1 dx=12ln⁡(x2+1)+C\int \frac{x}{x^2 + 1}\, dx = \frac{1}{2}\ln(x^2 + 1) + C, where the numerator is half the derivative of the denominator.

Why it happens: the pattern ∫f′(x)f(x) dx=ln⁡∣f(x)∣+C\int \frac{f'(x)}{f(x)}\, dx = \ln|f(x)| + C is remembered as "a fraction gives a logarithm", and the check on the numerator is skipped.

How to check: differentiate the answer. ddx[ln⁡(x2+1)]=2xx2+1\frac{d}{dx}[\ln(x^2 + 1)] = \frac{2x}{x^2 + 1}, which is 00 at x=0x = 0, while the integrand 1x2+1\frac{1}{x^2 + 1} is 11 there.

Special Integrals FAQ

What is the power rule for integration?

+
∫xⁿ dx = xⁿ⁺¹/(n+1) + C, valid for every n except −1. Raise the exponent by one, then divide by the new exponent — the exact reverse of the derivative power rule, which lowers the exponent and multiplies by the old one. The exclusion of n = −1 is not a technicality: dividing by n + 1 would divide by zero.Read more →

Why is ∫(1/x) dx = ln|x| instead of ln x?

+
Because 1/x is defined for negative x but ln x is not. For x > 0 the antiderivative is ln x; for x < 0 it is ln(−x), whose derivative is also 1/x by the chain rule. The absolute value packs both branches into one formula. Strictly, each branch carries its own constant, since the domain splits at zero.Read more →

What is the integral of a^x?

+
∫aˣ dx = aˣ/ln a + C, for a > 0 and a ≠ 1. The 1/ln a compensates for the chain rule: differentiating aˣ produces aˣ ln a, so the antiderivative has to divide that factor back out. When a = e the factor is ln e = 1, which is why ∫eˣ dx = eˣ + C carries no coefficient.Read more →

Why is ∫sin x dx = −cos x?

+
Because the derivative that produces sin x carries a minus sign. Since (cos x)′ = −sin x, reversing it means the antiderivative of sin x is −cos x, not cos x. Check by differentiating: (−cos x)′ = sin x. The same inversion explains ∫csc²x dx = −cot x, since (cot x)′ = −csc²x.Read more →

How do you spot an integral that gives a logarithm?

+
Look for the pattern f′(x)/f(x) — a function in the denominator with its own derivative on top. Any integrand of that shape integrates to ln|f(x)| + C. It shows up in disguise constantly: ∫tan x dx is ∫sin x / cos x dx, which is −f′/f with f = cos x, giving −ln|cos x| + C.Read more →

What is the integral of sec x?

+
∫sec x dx = ln|sec x + tan x| + C. It is the least obvious of the standard trigonometric integrals, because you get there by multiplying by a clever form of 1: sec x · (sec x + tan x)/(sec x + tan x). The numerator then becomes the derivative of the denominator, turning it into the f′/f pattern.Read more →