Chapter 5.10, Problem 49E

To determine

To find the value of the integral ${\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{1}{\sqrt{x}\text{(}1+x)}dx}$ by reducing it into improper integrals of Type 1 and Type 2.

Answer to Problem 49E

The value of the integral is $\pi$

Explanation of Solution

Given information: ${\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{1}{\sqrt{x}\text{(}1+x)}dx}={\displaystyle \underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{x}\text{(}1+x)}dx}+{\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{1}{\sqrt{x}\text{(}1+x)}dx}$

Formula used:

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

Calculation : To evaluate the integral of type 1 and type 2, the integral of the form ${\displaystyle \int \frac{1}{\sqrt{x}\text{(}1+x)}dx}$ needs to be calculated first.

Put $t^2=x$ so that $dx=2tdt$

The integral becomes as: ${\displaystyle \int \frac{2t}{t\text{(}1+t^2)}dt}$

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

The integral ${\displaystyle \underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{x}\text{(}1+x)}dx}=2{\left[{\tan }^{-1}x\right]}_0^1=2\left[{\tan }^{-1}(1)-{\tan }^{-1}(0)\right]=2.\frac{\pi }{4}=\frac{\pi }{2}$

and the integral ${\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{1}{\sqrt{x}\text{(}1+x)}dx}=2{\left[{\tan }^{-1}x\right]}_1^{\infty }=2\left[{\tan }^{-1}(\infty )-{\tan }^{-1}(1)\right]=2\left[\frac{\pi }{2}-\frac{\pi }{4}\right]=\frac{\pi }{2}$

  $\left[\text{since}{\tan }^{-1}(\infty )=\frac{\pi }{2},{\tan }^{-1}(1)=\frac{\pi }{4}\right]$

Thus, the integral ${\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{1}{\sqrt{x}\text{(}1+x)}dx}=\frac{\pi }{2}+\frac{\pi }{2}=\pi$

Therefore, the value of the integral is $\pi$