Chapter 6, Problem 42RE

To determine

To calculate: $\frac{dy}{dx}$ from $y={\displaystyle \underset{x}{\overset{2x}{\int }}\frac{1}{t^2+1}}dt$ .

Answer to Problem 42RE

  $\frac{-3x^2}{\left(1+x^2\right)\left(1+4x^2\right)}$

Explanation of Solution

Given information: The integral $y={\displaystyle \underset{x}{\overset{2x}{\int }}\frac{1}{t^2+1}}dt$

Formula used: Use integration formula ${\displaystyle \int \frac{1}{x^2+1}dx}=\tan x+c$ and differentiation formula $\frac{d}{dx}\left(\tan x\right)=\frac{1}{1+x^2}$

Calculation:

First of all we have to solve the integral $y={\displaystyle \underset{x}{\overset{2x}{\int }}\frac{1}{t^2+1}}dt$

  $\begin{array}{c}y={\displaystyle \underset{x}{\overset{2x}{\int }}\frac{1}{t^2+1}}dt\\ ={\left[\tan t\right]}_x^{2x}\\ =\tan 2x-\tan x\end{array}$

Hence, $y=\tan 2x-\tan x$

Now differentiate $y=\tan 2x-\tan x$ with respect to $x$

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left(\tan 2x-\tan x\right)\\ =\frac{d}{dx}\left(\tan 2x\right)-\frac{d}{dx}\left(\tan x\right)\\ =\frac{1}{1+{\left(2x\right)}^2}-\frac{1}{1+{\left(x\right)}^2}\\ =\frac{1}{1+4x^2}-\frac{1}{1+x^2}\end{array}$

Further simplify it

  $\begin{array}{c}\frac{1}{1+4x^2}-\frac{1}{1+x^2}=\frac{1-x^2-1-4x^2}{\left(1+x^2\right)\left(1+4x^2\right)}\\ =\frac{-3x^2}{\left(1+x^2\right)\left(1+4x^2\right)}\end{array}$

Therefore, $\frac{dy}{dx}=\frac{-3x^2}{\left(1+x^2\right)\left(1+4x^2\right)}$