MCQ
$\int_{}^{} {\frac{{2x{{\tan }^{ - 1}}{x^2}}}{{1 + {x^4}}}} \;dx = $
- A${[{\tan ^{ - 1}}{x^2}]^2} + c$
- ✓$\frac{1}{2}{[{\tan ^{ - 1}}{x^2}]^2} + c$
- C$2{[{\tan ^{ - 1}}{x^2}]^2} + c$
- DNone of these
✓
Answer
Correct option: B.
$\frac{1}{2}{[{\tan ^{ - 1}}{x^2}]^2} + c$
b
(b) Put $t = {\tan ^{ - 1}}{x^2} \Rightarrow dt = \frac{1}{{1 + {x^4}}}\,2x\,dx,$ then
$\int_{}^{} {\frac{{2x{{\tan }^{ - 1}}{x^2}}}{{1 + {x^4}}}} \,dx = \int_{}^{} {t\,dt} = \frac{{{t^2}}}{2} + c = \frac{1}{2}{({\tan ^{ - 1}}{x^2})^2} + c.$
(b) Put $t = {\tan ^{ - 1}}{x^2} \Rightarrow dt = \frac{1}{{1 + {x^4}}}\,2x\,dx,$ then
$\int_{}^{} {\frac{{2x{{\tan }^{ - 1}}{x^2}}}{{1 + {x^4}}}} \,dx = \int_{}^{} {t\,dt} = \frac{{{t^2}}}{2} + c = \frac{1}{2}{({\tan ^{ - 1}}{x^2})^2} + c.$
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