$\smallint \frac{{{x^2}}}{{{x^2} + \,\,4}}\,\;dx$ is equal to

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

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

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

\int \frac{{{x^2}}}{{{x^2} + \,\,4}}\,\;dx

is equal to

View answer and explanation

Correct option: A

x\,\, - \,\,2\,{\tan ^{ - 1}}\,\,\left( {\frac{x}{2}} \right) + c

Option A (

x - 2{\tan ^{ - 1}}(\frac{x}{2}) + c

) is correct because after dividing and using the arctangent integral, this is the result.