$\int \frac{(\tan^{-1} x)^3}{1 + x^2} \, dx = $
- A$\frac{(\tan^{-1} x)^4}{4} + c$
- B$\frac{(\tan^{-1} x)^4}{4} + c$
- C$2 \tan^{-1} x + c$
- D$2 (\tan^{-1} x)^2 + c$
Explore More
Similar Questions
$\int \frac{x^2}{1+x^6} d x$ is equal to
Integrate the function $\cot x \log \sin x$.
Easy
View SolutionThe value of $\int \frac{dx}{x^2(x^4+1)^{3/4}}$ is
Integrate the function: $\frac{\sec ^{2} x}{\sqrt{\tan ^{2} x+4}}$
Easy
View SolutionLet $f(x) = \int x^3 \sqrt{3-x^2} dx$. If $5f(\sqrt{2}) = -4$,then $f(1)$ is equal to
DifficultJEE MAIN 2025
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo