Question

$\int \frac{(x^2 -1)}{(x^2 + 1) \sqrt{x^4 + 1}} dx $ is equal to

• A. $\sec^{-1} \left(\frac{x^{2} +1}{\sqrt{2}x}\right) + c $
• B. $\frac{1}{\sqrt{2}} \sec^{-1} \left(\frac{x^{2} + 1}{ \sqrt{2} x}\right)+c $
• C. $\frac{1}{\sqrt{2}} \sec^{-1} \left(\frac{x^{2} + 1}{ \sqrt{2}}\right)+c $
• D. None of these

Answer 1

Answer: (B)

$I=\int \frac{x^{2}\left(1-\frac{1}{x^{2}}\right) d x}{x^{2}\left(x+\frac{1}{x}\right)\left(x^{2}+\frac{1}{x^{2}}\right)^{1 / 2}}$
Let $x+\frac{1}{x}=p$
$\Rightarrow \left(1-\frac{1}{x^{2}}\right) d x=d p$
$I=\int \frac{d p}{p \sqrt{p^{2}-2}}=\frac{1}{\sqrt{2}} \sec ^{-1} \frac{p}{\sqrt{2}}$
$=\frac{1}{\sqrt{2}} \sec ^{-1} \frac{p}{\sqrt{2}}$
$=\frac{1}{\sqrt{2}} \sec ^{-1}\left(\frac{x^{2}+1}{\sqrt{2} x}\right)+c$