Question

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

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

Answer 1

Answer: (A)

$\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{x^2-1}{\sqrt{2}x}\right)+C$

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

$=\int \frac{x^2(1+\frac{1}{x^2})}{x^2(x^2+\frac{1}{x^2})}dx$

$=\int \frac{1+\frac{1}{x^2}}{(x-\frac{1}{x}{)}^2+2}dx$

Take $x-\frac{1}{x}=t$ $\Rightarrow (1+\frac{1}{x^2})dx=dt$

$I=\int \frac{dt}{t^2+2}$

$=\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{t}{\sqrt{2}}+C$

$=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{x^2-1}{\sqrt{2}x}\right)+C$