Question

Evaluate : $\int {\cot }^{-1}(1+x+x^2)dx$

Answer 1

${\text{cot}}^{-1}(1+x+x^2)={\text{tan}}^{-1}(\frac{x+1-x}{1+x(x+1)})={\text{tan}}^{-1}(x+1)-{\text{tan}}^{-1}\left(x\right)$

$\int {\text{cot}}^{-1}(1+x+x^2)dx=\int 1.{\text{tan}}^{-1}(x+1)dx-\int 1.{\text{tan}}^{-1}xdx$

$=x{\text{tan}}^{-1}(x+1)-\int \frac{xdx}{(x+1{)}^2+1}-x{\text{tan}}^{-1}x+\int \frac{xdx}{x^2+1}$

$=x\left({\text{tan}}^{-1}\right(x+1)-{\tan }^{-1}x)-\frac{1}{2}\ln \left[\right(x+1{)}^2+1]+\frac{1}{2}\ln (x^2+1)$

$=x{\text{cot}}^{-1}(1+x+x^2)-\frac{1}{2}\ln |\frac{(x+1{)}^2+1}{x^2+1}|+C$