Question

$\int \frac{x^2+1}{x^4+1}dx$

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

Answer 1

Answer: (D)

$\int \frac{x^2+1}{x^4+1}dx=\int \frac{\left(1+\frac{1}{x^2}\right)}{\left(x^2+\frac{1}{x^2}\right)}dx$
$=\int \frac{\left(1+\frac{1}{x^2}\right)}{{\left(x-\frac{1}{x}\right)}^2+2}dx$
Put $x-\frac{1}{x}=t\Rightarrow \left(1+\frac{1}{x^2}\right)dx=dt$
$=\int \frac{dt}{t^2+{\left(\sqrt{2}\right)}^2}=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{t}{\sqrt{2}}\right)+c$
$=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{x^2-1}{\sqrt{2}x}\right)+c$