Question

If $\int x\log (1+x^2)dx=\varphi \left(x\right).\log (1+x^2)+\Psi \left(x\right)+c$ then

• A. $\varphi \left(x\right)=\frac{1+x^2}{2}$
• B. $\Psi \left(x\right)=\frac{1+x^2}{2}$
• C. $\Psi \left(x\right)=-\frac{1+x^2}{2}$
• D. $\varphi \left(x\right)=-\frac{1+x^2}{2}$

Answer 1

Answer: (A), (C)

$\varphi \left(x\right)=\frac{1+x^2}{2}$
$\Psi \left(x\right)=-\frac{1+x^2}{2}$

Given, $\int x\log (1+x^2)dx=\varphi \left(x\right).\log (1+x^2)+\Psi \left(x\right)+c$
Consider, $\int x\log (1+x^2)dx$
$=\log (1+x^2)\frac{x^2}{2}-\int (\frac{2x}{1+x^2})\frac{x^2}{2}dx$
$=\log (1+x^2)\frac{x^2}{2}-\int \frac{x^3}{1+x^2}dx$
$=\frac{x^2}{2}\log (1+x^2)+\int (-x+\frac{x}{x^2+1})dx$
$=\frac{x^2}{2}\log (1+x^2)-\frac{x^2}{2}+\int \frac{x}{x^2+1}dx$
Substitute $x^2+1=t$
$\Rightarrow xdx=\frac{dt}{2}$
$=\frac{x^2}{2}\log (1+x^2)-\frac{x^2}{2}+\frac{1}{2}\int \frac{1}{t}dt$
$=\frac{x^2}{2}\log (1+x^2)-\frac{x^2}{2}+\frac{1}{2}\log (1+x^2)+C$
$=\frac{x^2}{2}\log (1+x^2)-\frac{x^2}{2}+\frac{1}{2}\log (1+x^2)+C$
$=\frac{x^2+1}{2}\log (1+x^2)-\frac{x^2+1}{2}+C^{\prime }$
On comparing with (1), we get
$\varphi \left(x\right)=\frac{x^2+1}{2},\psi \left(x\right)=-\frac{x^2+1}{2}$