Question

Integrate $\frac{1}{2}f{}^{\prime }\left(x\right)w.r.t.x^4$, where $f\left(x\right)=ta{n}^{-1}x+ln\sqrt{1+x}-ln\sqrt{1-x}$

• A. $-ln(1-x^4)+C$
• B. $-ln(1+x^2)+C$
• C. $-ln(1-x^2)+C$
• D. $-ln(1+x^4)+C$

Answer 1

Answer: (A)

$-ln(1-x^4)+C$

$f\left(x\right)=ta{n}^{-1}x+ln\sqrt{1+x}-ln\sqrt{1-x}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{1+x^2}+\frac{1}{2}[\frac{1}{1+x}+\frac{1}{1-x}]$

$=\frac{1}{1+x^2}+\frac{1}{1-x^2}=\frac{2}{1-x^4}$

Now
$\int \frac{1}{2}f{}^{\prime }\left(x\right)d\left(x^4\right)=\int \frac{1}{2}\left(\frac{2}{1-x^4}\right)d\left(x^4\right)$

$=\int \frac{4x^3}{1-x^4}dx=-ln(1-x^4)+C$