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Indefinite IntegrationIntegration as anti-derivative
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After the notes

Formula sheet

30 results from this chapter, grouped the way the notes teach them.

Standard integrals

∫xn dx\int x^n\,dx
xn+1n+1,  n≠−1\dfrac{x^{n+1}}{n+1},\ \ n\ne-1
∫dxx\int \dfrac{dx}{x}
ln⁡∣x∣\ln|x|
∫ex dx,  ∫ax dx\int e^x\,dx,\ \ \int a^x\,dx
ex,  axln⁡ae^x,\ \ \dfrac{a^x}{\ln a}
∫tan⁡x dx,  ∫cot⁡x dx\int \tan x\,dx,\ \ \int \cot x\,dx
ln⁡∣sec⁡x∣,  ln⁡∣sin⁡x∣\ln|\sec x|,\ \ \ln|\sin x|
∫sec⁡x dx\int \sec x\,dx
ln⁡∣sec⁡x+tan⁡x∣\ln|\sec x+\tan x|
∫csc⁡x dx\int \csc x\,dx
ln⁡∣csc⁡x−cot⁡x∣\ln|\csc x-\cot x|
∫f(ax+b) dx\int f(ax+b)\,dx, if ∫f(x) dx=F(x)\int f(x)\,dx=F(x)
1aF(ax+b)\dfrac1a F(ax+b)

Quadratic forms

∫dxx2+a2\int \dfrac{dx}{x^2+a^2}
1atan⁡−1xa\dfrac1a\tan^{-1}\dfrac xa
∫dxx2−a2\int \dfrac{dx}{x^2-a^2}
12aln⁡∣x−ax+a∣\dfrac1{2a}\ln\left|\dfrac{x-a}{x+a}\right|
∫dxa2−x2\int \dfrac{dx}{a^2-x^2}
12aln⁡∣a+xa−x∣\dfrac1{2a}\ln\left|\dfrac{a+x}{a-x}\right|
∫dxa2−x2\int \dfrac{dx}{\sqrt{a^2-x^2}}
sin⁡−1xa\sin^{-1}\dfrac xa
∫dxxx2−a2\int \dfrac{dx}{x\sqrt{x^2-a^2}}
1asec⁡−1xa\dfrac1a\sec^{-1}\dfrac xa
∫dxx2±a2\int \dfrac{dx}{\sqrt{x^2\pm a^2}}
ln⁡∣x+x2±a2∣\ln\left|x+\sqrt{x^2\pm a^2}\right|
∫a2−x2 dx\int \sqrt{a^2-x^2}\,dx
x2a2−x2+a22sin⁡−1xa\dfrac x2\sqrt{a^2-x^2}+\dfrac{a^2}2\sin^{-1}\dfrac xa
∫x2±a2 dx\int \sqrt{x^2\pm a^2}\,dx
x2x2±a2±a22ln⁡∣x+x2±a2∣\dfrac x2\sqrt{x^2\pm a^2}\pm\dfrac{a^2}2\ln\left|x+\sqrt{x^2\pm a^2}\right|
∫dxquadratic\int\dfrac{dx}{\text{quadratic}}: discriminant D<0D<0 / D>0D>0
tan⁡−1\tan^{-1} form / ln⁡\ln form (complete the square)

Substitution and by parts

a2−x2\sqrt{a^2-x^2}, a2+x2\sqrt{a^2+x^2}, x2−a2\sqrt{x^2-a^2}
x=asin⁡θx=a\sin\theta, x=atan⁡θx=a\tan\theta, x=asec⁡θx=a\sec\theta
a2−x2a2+x2\sqrt{\dfrac{a^2-x^2}{a^2+x^2}}
x2=a2cos⁡2θx^2=a^2\cos2\theta
∫f′(x)f(x) dx\int \dfrac{f'(x)}{f(x)}\,dx
ln⁡∣f(x)∣\ln|f(x)|
∫[f(x)]nf′(x) dx\int [f(x)]^n f'(x)\,dx
[f(x)]n+1n+1,  n≠−1\dfrac{[f(x)]^{n+1}}{n+1},\ \ n\ne-1
Integration by parts
∫uv dx=u∫v dx−∫(u′∫v dx)dx\int uv\,dx=u\int v\,dx-\int\left(u'\int v\,dx\right)dx
Choosing the first function
ILATE: Inverse trig, Log, Algebraic, Trig, Exponential
∫ex(f(x)+f′(x)) dx\int e^x\big(f(x)+f'(x)\big)\,dx
exf(x)e^x f(x)
∫(f(x)+xf′(x)) dx\int \big(f(x)+xf'(x)\big)\,dx
xf(x)xf(x)
∫ex p(x) dx\int e^x\,p(x)\,dx, pp a polynomial of degree nn
exq(x)e^x q(x), qq of degree nn: differentiate and equate coefficients

Trigonometric types

∫dxa+bsin⁡2x\int\dfrac{dx}{a+b\sin^2x}, dxasin⁡2x+bcos⁡2x+c\dfrac{dx}{a\sin^2x+b\cos^2x+c}
Multiply num. and den. by sec⁡2x\sec^2x, put tan⁡x=t\tan x=t
∫dxa+bsin⁡x\int\dfrac{dx}{a+b\sin x}, dxasin⁡x+bcos⁡x+c\dfrac{dx}{a\sin x+b\cos x+c}
Put tan⁡x2=t\tan\dfrac x2=t
∫asin⁡x+bcos⁡x+clsin⁡x+mcos⁡x+n dx\int\dfrac{a\sin x+b\cos x+c}{l\sin x+m\cos x+n}\,dx
Num. =A⋅=A\cdotDen. +B⋅ddx+B\cdot\dfrac{d}{dx}(Den.) +C+C
∫x2±1x4+kx2+1 dx\int\dfrac{x^2\pm1}{x^4+kx^2+1}\,dx
Divide by x2x^2; put x∓1x=tx\mp\dfrac1x=t
sin⁡x±cos⁡x\sin x\pm\cos x in the numerator, sin⁡2x\sin2x in the denominator
Write sin⁡2x=(sin⁡x+cos⁡x)2−1\sin2x=(\sin x+\cos x)^2-1 or 1−(sin⁡x−cos⁡x)21-(\sin x-\cos x)^2
Last look

Quick revision

The checks to run before you sit a question on functions.

  1. 1Simplify first: try to bring the integrand into, or very close to, one of the standard forms.
  2. 2Never forget +C+C, and keep ∣ ∣|\ | inside every ln⁡\ln.
  3. 3If the derivative of one part sits next to it, substitute that part — ∫f(g(x)) g′(x) dx\int f(g(x))\,g'(x)\,dx.
  4. 4For tan⁡10xtan⁡7xtan⁡3x\tan10x\tan7x\tan3x-type products, use tan⁡Atan⁡Btan⁡(A+B)=tan⁡(A+B)−tan⁡A−tan⁡B\tan A\tan B\tan(A+B)=\tan(A+B)-\tan A-\tan B.
  5. 5By parts: pick the first function with ILATE; when only one function is visible, take 11 as the second.
  6. 6Seeing exe^x times a sum? Check whether it splits as f+f′f+f' before doing by parts.
  7. 7Partial fractions need a proper fraction — divide first if the degree of the numerator is not lower.
  8. 8Linear over quadratic: write the numerator as A⋅ddx(den.)+BA\cdot\frac{d}{dx}(\text{den.})+B.
  9. 9If an integral is said to be rational, the coefficients of the 1x−a\frac1{x-a} terms in its partial fractions must be 00.
  10. 10For ∫tan⁡x\int\sqrt{\tan x} or cot⁡x\sqrt{\cot x}, put tan⁡x=y2\tan x=y^2 and split into y2+1y4+1\frac{y^2+1}{y^4+1} and y2−1y4+1\frac{y^2-1}{y^4+1}.