MCQ
$\int_{}^{} {\frac{{3{x^2}}}{{{x^6} + 1}}dx = } $
- A$\log ({x^6} + 1) + c$
- ✓${\tan ^{ - 1}}({x^3}) + c$
- C$3{\tan ^{ - 1}}({x^3}) + c$
- D$3{\tan ^{ - 1}}\left( {\frac{{{x^3}}}{3}} \right) + c$
✓
Answer
Correct option: B.
${\tan ^{ - 1}}({x^3}) + c$
b
(b) Put ${x^3} = t \Rightarrow 3{x^2}dx = dt,$ therefore
$\int_{}^{} {\frac{{3{x^2}}}{{{x^6} + 1}}\,dx = \int_{}^{} {\frac{1}{{{t^2} + 1}}dt = {{\tan }^{ - 1}}(t) + c} } = {\tan ^{ - 1}}({x^3}) + c$.
(b) Put ${x^3} = t \Rightarrow 3{x^2}dx = dt,$ therefore
$\int_{}^{} {\frac{{3{x^2}}}{{{x^6} + 1}}\,dx = \int_{}^{} {\frac{1}{{{t^2} + 1}}dt = {{\tan }^{ - 1}}(t) + c} } = {\tan ^{ - 1}}({x^3}) + c$.
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