How do you integrate int 1/[(1+x)(1-2x)] using partial fractions?

Mar 16, 2016

$\int \frac{1}{\left(1 + x\right) \left(1 - 2 x\right)} \mathrm{dx}$

= $\frac{1}{3} \ln \left(1 + x\right) - \frac{1}{3} \ln \left(1 - 2 x\right) + c$

Explanation:

To integrate, we should first convert $\frac{1}{\left(1 + x\right) \left(1 - 2 x\right)}$ in to partial fractions. Let

$\frac{1}{\left(1 + x\right) \left(1 - 2 x\right)} \Leftrightarrow \frac{A}{1 + x} + \frac{B}{1 - 2 x}$. Simplifying RHS

=[A(1-2x)+B(1+x)]/((1+x)(1-2x) or

=[A-2Ax+B+Bx]/((1+x)(1-2x)

=$\frac{\left(B - 2 A\right) x + \left(A + B\right)}{\left(x + 3\right) \left(x - 7\right) \left(x + 4\right)}$

Hence $B - 2 A = 0$, $A + B = 1$ or $A = 1 - B$

Hence $B - 2 \left(1 - B\right) = 0$ or $3 B = 2$ or

$B = \frac{2}{3}$ and hence $A = \frac{1}{3}$

Hence $\frac{1}{\left(1 + x\right) \left(1 - 2 x\right)} \Leftrightarrow \frac{1}{3 \left(1 + x\right)} + \frac{2}{3 \left(1 - 2 x\right)}$

Hence $\int \frac{1}{\left(1 + x\right) \left(1 - 2 x\right)} \mathrm{dx}$ =

$\int \left[\frac{1}{3 \left(1 + x\right)} + \frac{2}{3 \left(1 - 2 x\right)}\right] \mathrm{dx}$

Now one can use the identity $\int \left(\frac{1}{a x + b}\right) \mathrm{dx} = \frac{1}{a} \ln \left(a x + b\right)$

Hence, $\int \left[\frac{1}{3 \left(1 + x\right)} + \frac{2}{3 \left(1 - 2 x\right)}\right] \mathrm{dx}$

= $\frac{1}{3} \ln \left(1 + x\right) + \frac{2}{3} \times \left(- \frac{1}{2}\right) \ln \left(1 - 2 x\right) + c$

= $\frac{1}{3} \ln \left(1 + x\right) - \frac{1}{3} \ln \left(1 - 2 x\right) + c$