Mathematics

$\displaystyle\int_{0}^{1}\dfrac{1-x^{2}}{(1+x^{2})^{2}}dx$

SOLUTION

$I=\displaystyle\int_{0}^{1}\dfrac{1-x^{2}}{(1+x^{2})^{2}}dx$
$I=\displaystyle\int_{0}^{1}\dfrac{{\dfrac{1}{x^2}-1}}{\left(x+\dfrac{1}{x}\right)^{2}}\ dx$

Let $x+\dfrac{1}{x}=t$.

Then, $d\left(x+\dfrac{1}{x}\right)=dt$

$\Rightarrow \left(1-\dfrac{1}{x^{2}}\right)dx=dt$

Clearly,

$x=0\Rightarrow t=\infty$

$x=1\Rightarrow t=2$

$\therefore =\displaystyle\int_{\infty}^{2}-\dfrac{1}{t^{2}}dt$

$\left[\dfrac{1}{t}\right]_{\infty}^{2}=\dfrac{1}{2}$

Its FREE, you're just one step away

Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

Realted Questions

Q1 Subjective Hard
$\int {\dfrac {(x+1)}{x(1+xe^{x})^{2}}}dx$ is equal to

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Solve:
$\int {{{\cos }^{ - 1}}x} \,dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Find the integral:
$\int {\cos \sqrt x dx\,.}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
If $\int _{ 0 }^{ 100 }{ f\left( x \right) dx=a }$, then $\sum _{ r=1 }^{ 100 }{ \int _{ 0 }^{ 1 }{ (f\left( r-1+x \right) dx) } }$=
• A. $100a$
• B. $0$
• C. $10a$
• D. $a$

If $\displaystyle \int \dfrac{e^x - 1}{e^x + 1}dx =f(x) + c$ then $f(x) =$