Question

Prove that ${\int }_1^2\frac{dx}{x(x+1{)}^2}dx=\log \left(\frac{4}{3}\right)+\frac{1}{6}$.

Answer 1

Let $\frac{1}{x{(x+1)}^2}=\frac{A}{x+1}+\frac{B}{{(x+1)}^2}+\frac{C}{x}$
$\Rightarrow 1=Ax(x+1)+Bx+C{(x+1)}^2$
Comparing coefficients and solving , we get
$A=-1,B=-1,C=1$
Hence
$I={\int }_1^2\frac{dx}{x{(x+1)}^2}={\int }_1^2(-\frac{1}{x+1}-\frac{1}{{(x+1)}^2}+\frac{1}{x})dx$
$={[-\log (x+1)+\frac{1}{x+1}+\log x]}_1^2$

$\Rightarrow I=\log \frac{4}{3}+\frac{1}{6}$