Question

${\int }_0^a\frac{1}{a^2+x^2}d{x}_{=}$

• A. $\pi /2$
• B. $\pi /3$
• C. $\pi /4$
• D. $\pi /4a$

Answer 1

The correct option is D $\pi /4a$

${\int }_0^a\frac{1}{a^2+x^2}dx$

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

$\int \frac{d\left(\frac{x}{a}\right)}{1+(\frac{x}{a}{)}^2}=\frac{1}{a}\int \frac{d\left(\frac{x}{a}\right)}{1+(\frac{x}{a}{)}^2}$

$=\frac{1}{a}ta{n}^{-1}\left(\frac{x}{a}\right)+c$

${\int }_0^a\frac{1}{a^2+x^2}dx=[\frac{1}{a}ta{n}^{-1}\left(\frac{x}{a}\right)+c]{\int }_0^a$

$=[\frac{1}{a}ta{n}^{-1}\left(\frac{a}{a}\right)+c]-(0+c)$
$=\frac{1}{a}ta{n}^{-1}\left(1\right)$

$=\frac{1}{a}\cdot \frac{\pi }{4}$

$=\frac{\pi }{4a}$