Question

${\displaystyle \int \frac{x^2+1}{x^4-x^2+1}}$dx = ?

विकल्प

• ${\displaystyle {\tan }^{-1}\left(\frac{x^2+1}{2}\right)+\text{c}}$

• tan^-1 (x²) + c

• tan^-1 (2x² - 1) + c

• ${\displaystyle {\tan }^{-1}\left(\frac{x^2-1}{x}\right)+\text{c}}$

Answer 1

${\displaystyle {\tan }^{-1}\left(\frac{x^2-1}{x}\right)+\text{c}}$

Explanation:

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

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

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

put ${\displaystyle x-\frac{1}{x}}$ = t

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

${\displaystyle \therefore \text{I}=\int \frac{\text{dt}}{{\text{t}}^2+1}={\tan }^{-1}\left(\text{t}\right)+\text{c}}$

${\displaystyle ={\tan }^{-1}(x-\frac{1}{x})+\text{c}}$

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