# int_0^1 1/(x^2+1) = ?

Dec 14, 2017

$\frac{\pi}{4}$

#### Explanation:

First thing to sonsider is:

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

Make a substitution:

$x = \tan \theta$

$\implies \mathrm{dx} = {\sec}^{2} \theta$ $d \theta$

$\implies \int \frac{1}{{\tan}^{2} \theta + 1} {\sec}^{2}$$d \theta$

We must use the identity:

$1 + {\tan}^{2} x = {\sec}^{2} x$

$\implies \int {\sec}^{2} \frac{\theta}{{\sec}^{2} \theta} d \theta$

$\implies \int d \theta$

$\implies \theta + c$

Use: ${\tan}^{- 1} x = \theta$

$= {\tan}^{- 1} x + c$

Now considering limits from $0$ to $1$

$\implies {\tan}^{- 1} \left(1\right) - {\tan}^{- 1} \left(0\right)$

$\implies \frac{\pi}{4} - 0$

$\implies \frac{\pi}{4}$

Dec 14, 2017

$\frac{\pi}{4}$

#### Explanation:

${\int}_{0}^{1} \frac{1}{{x}^{2} + 1} \mathrm{dx}$

we do this by substitution

$x = \tan u$

$\implies \mathrm{dx} = {\sec}^{2} u \mathrm{du}$

change of limits

$x = 0 \implies u = {\tan}^{- 1} 0 = 0$

$x = 1 \implies u = {\tan}^{- 1} 1 = \frac{\pi}{4}$

the integral becomes

${\int}_{0}^{\frac{\pi}{4}} \frac{1}{\cancel{{\tan}^{2} u + 1}} \times \cancel{{\sec}^{2} u} \mathrm{du}$

$= {\int}_{0}^{\frac{\pi}{4}} \mathrm{du} = {\left[u\right]}_{0}^{\frac{\pi}{4}}$

$= \frac{\pi}{4} = 0 = \frac{\pi}{4}$