Question:
\(\int \dfrac{x\tan^{-1} x}{(1-x^2)^{3/2}}\,dx = ?\)
\(\int \dfrac{x\tan^{-1} x}{(1-x^2)^{3/2}}\,dx\) किसके बराबर है?
Options:
- \(\dfrac{x\tan^{-1} x}{\sqrt{1+x^2}} + \dfrac{1}{2}\log\left|\dfrac{x}{\sqrt{1+x^2}}\right| + C\)
- \(\dfrac{\tan^{-1} x}{\sqrt{1+x^2}} – \dfrac{x}{\sqrt{1+x^2}} + C\)
- \(\dfrac{-\tan^{-1} x}{\sqrt{1+x^2}} + \dfrac{x}{\sqrt{1+x^2}} + C\)
- \(\dfrac{x\tan^{-1} x}{\sqrt{1-x^2}} – \dfrac{1}{2}\log\left|\dfrac{x}{\sqrt{1-x^2}}\right| + C\)
- \(\dfrac{x\tan^{-1} x}{\sqrt{1+x^2}} + \dfrac{1}{2}\log\left|\dfrac{x}{\sqrt{1+x^2}}\right| + C\)
- \(\dfrac{\tan^{-1} x}{\sqrt{1+x^2}} – \dfrac{x}{\sqrt{1+x^2}} + C\)
- \(\dfrac{-\tan^{-1} x}{\sqrt{1+x^2}} + \dfrac{x}{\sqrt{1+x^2}} + C\)
- \(\dfrac{x\tan^{-1} x}{\sqrt{1-x^2}} – \dfrac{1}{2}\log\left|\dfrac{x}{\sqrt{1-x^2}}\right| + C\)
Correct Answer: c)
Explanation:
Using substitution and integration by parts yields:
\(\int \dfrac{x\tan^{-1} x}{(1-x^2)^{3/2}}\,dx = \dfrac{-\tan^{-1} x}{\sqrt{1+x^2}} + \dfrac{x}{\sqrt{1+x^2}} + C\).
खंडशः समाकलन (integration by parts) का प्रयोग करने पर विकल्प (c) प्राप्त होता है।