Question

Find $\int \frac{dx}{x{(x^3+1)}^2}$

Answer 1

Consider the given function:

$I=\int \frac{1}{x{(x^3+1)}^2}dx$

$=\int \frac{x^2}{x^3{(x^3+1)}^2}dx$

$let{x}^3=t$

$3x^{2}dx=dx$

$=\frac{1}{3}\int \frac{1}{t{(t+1)}^2}dt$

$weknowthat$

$\frac{1}{t{(t+1)}^2}=\frac{A}{t}+\frac{B}{t+1}+\frac{C}{{(t+1)}^2}$

$\frac{1}{t{(t+1)}^2}=\frac{A{(t+1)}^2+B(t+1)+C\left(t\right)}{t{(t+1)}^2}$

$1=A{(t+1)}^2+B(t+1)+C\left(t\right)$

$putt=-1$

$1=A.0+B.(-1).0+C(-1)=-1$

$putt=0$

$1=A+B.\left(0\right)+C.\left(0\right)=1$

Hence this is the answer.