Integration Techniques

Integrate by substitution, by parts, and using partial fractions at A-Level.

A-Level integration goes beyond basic reverse differentiation to include substitution, integration by parts, and partial fractions.

Integration by Substitution

Replace a complex expression with uu.

2x(x2+1)3dx\int 2x(x^2+1)^3 dx. Let u=x2+1u = x^2+1, du=2xdxdu = 2x\,dx.

=u3du=u44+C=(x2+1)44+C= \int u^3\,du = \frac{u^4}{4} + C = \frac{(x^2+1)^4}{4} + C.

Integration by Parts

udv=uvvdu\int u\,dv = uv - \int v\,du

xexdx\int x e^x\,dx. Let u=xu = x, dv=exdxdv = e^x\,dx. Then du=dxdu = dx, v=exv = e^x.

=xexexdx=xexex+C=ex(x1)+C= xe^x - \int e^x\,dx = xe^x - e^x + C = e^x(x-1) + C.

Using Partial Fractions

5x+1(x+1)(x2)dx=2x+1+3x2dx=2lnx+1+3lnx2+C\int \frac{5x+1}{(x+1)(x-2)}\,dx = \int \frac{2}{x+1} + \frac{3}{x-2}\,dx = 2\ln|x+1| + 3\ln|x-2| + C.

Standard Integrals

f(x)f(x) f(x)dx\int f(x)\,dx
sinx\sin x cosx+C-\cos x + C
cosx\cos x sinx+C\sin x + C
sec2x\sec^2 x tanx+C\tan x + C
exe^x ex+Ce^x + C
1x\frac{1}{x} $\ln

Trig Integration

Use identities: sin2x=1cos2x2\sin^2 x = \frac{1-\cos 2x}{2}, cos2x=1+cos2x2\cos^2 x = \frac{1+\cos 2x}{2}.

Worked Example

xsinxdx\int x\sin x\,dx: by parts with u=xu = x, dv=sinxdxdv = \sin x\,dx.

=xcosx+cosxdx=xcosx+sinx+C= -x\cos x + \int \cos x\,dx = -x\cos x + \sin x + C.

Practice Problems

    1. (2x+1)5dx\int (2x+1)^5\,dx (substitution).
    1. xlnxdx\int x\ln x\,dx (by parts).
    1. 3(x+1)(x+2)dx\int \frac{3}{(x+1)(x+2)}\,dx (partial fractions).

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Key Takeaways

  • Substitution: simplify by replacing a sub-expression.

  • By parts: for products, especially x×x \times trig/exp.

  • Partial fractions: for rational functions.

  • Use trig identities for powers of sin/cos.

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