Question

Integrate $\int \frac{x(x^2+3)+3\left({\tan }^{-1}x\right){(1+x^2)}^2}{(1+x^2)}dx$ (where $C$ is an integration constant)

• A. $\frac{(x^2+3)}{(1+x^2)}{\tan }^{-1}x+C$
• B. $x(x^2+3){\tan }^{-1}x+C$
• C. $\frac{{\tan }^{-1}x}{1+x^2}+C$
• D. none

Answer 1

Answer: (B)

$x(x^2+3){\tan }^{-1}x+C$

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

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

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

$=\frac{x^2}{2}+\ln (1+x^2)+3\int (1+x^2){\tan }^{-1}xdx$

$=\frac{x^2}{2}+\ln (1+x^2)+3[(x+\frac{x^3}{3}){\tan }^{-1}x-\int \frac{x(3+x^2)}{3(1+x^2)}dx]+c$

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

$=(x^3+3x){\tan }^{-1}x+c$

$=x(x^2+3){\tan }^{-1}x+c$

Hence, the answer is $x(x^2+3){\tan }^{-1}x+c.$