Question

If $\int \log (a^2+x^2)dx=h\left(x\right),$ then $h\left(x\right)=$

• A. $x\log (a^2+x^2)+2{\tan }^{-1}\left(\frac{x}{a}\right)+c$
• B. $x^2\log (a^2+x^2)+x+a{\tan }^{-1}\left(\frac{x}{a}\right)+c$
• C. $x\log (a^2+x^2)-2x+2a{\tan }^{-1}\left(\frac{x}{a}\right)+c$
• D. $x^2\log (a^2+x^2)+2x-a^2{\tan }^{-1}\left(\frac{x}{a}\right)+c$

Answer 1

The correct option is C $x\log (a^2+x^2)-2x+2a{\tan }^{-1}\left(\frac{x}{a}\right)+c$

$I=\int \log (a^2+x^2)dx$

Integrating by parts, we get

Let $u=\log (a^2+x^2)\Rightarrow du=\frac{2xdx}{a^2+x^2}$

$dv=dx\Rightarrow v=x$

$I=x\log (a^2+x^2)-\int x\times \frac{2xdx}{a^2+x^2}$

$=x\log (a^2+x^2)-2\int \frac{x^{2}dx}{a^2+x^2}$

$=x\log (a^2+x^2)-2\int \frac{(x^2+a^2-a^2)dx}{a^2+x^2}$

$=x\log (a^2+x^2)-2\int \frac{(x^2+a^2)dx}{a^2+x^2}+2\int \frac{a^{2}dx}{a^2+x^2}$

$=x\log (a^2+x^2)-2\int dx+2a^2\int \frac{dx}{a^2+x^2}$

$=x\log (a^2+x^2)-2x+2a^2\times \frac{1}{a}{\tan }^{-1}\frac{x}{a}+c$

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