$x > 0$ के लिए $\int x \operatorname{Cos}^{-1}\left(\frac{1-x^2}{1+x^2}\right) d x$ का मान ज्ञात कीजिए।

  • A
    $-x + (1+x^2) \operatorname{Tan}^{-1} x + c$
  • B
    $x - (1+x^2) \operatorname{Cot}^{-1} x + c$
  • C
    $-x + (1+x^2) \operatorname{Cot}^{-1} x + c$
  • D
    $x - (1+x^2) \operatorname{Tan}^{-1} x + c$

Explore More

Similar Questions

समाकलन ज्ञात कीजिए: $\int \operatorname{Tan}^{-1}\left(x^{\frac{1}{3}}\right) d x$

$\int \cot x \cdot \log [\log (\sin x)] d x=$

$\int (\log_{e} 2x)^3 dx =$

$I = \int \cos(\ln x) \, dx$. तो $I =$

समाकलन ज्ञात कीजिए: $\int \frac{\sqrt{x^2+1}\left[\log \left(x^2+1\right)-2 \log x\right]}{x^4} d x$

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo