MCQ
$\int_{}^{} {{x^2}{{(3)}^{{x^3} + 1}}dx = } $
- A${(3)^{{x^3}}} + c$
- ✓$\frac{{{{(3)}^{{x^3}}}}}{{\log 3}} + c$
- C$\log 3{(3)^{{x^3}}} + c$
- DNone of these
✓
Answer
Correct option: B.
$\frac{{{{(3)}^{{x^3}}}}}{{\log 3}} + c$
b
(b)$\int_{}^{} {{x^2}{{(3)}^{{x^3} + 1}}dx} = \int_{}^{} {3{x^2}.\,{{(3)}^{{x^3}}}dx} $
Putting ${x^3} = t \Rightarrow 3{x^2}\,dx = dt,$ we get
$ = \int_{}^{} {{3^t}dt = \frac{{{3^t}}}{{\log 3}} + c = \frac{{{{(3)}^{{x^3}}}}}{{\log 3}} + c} $.
(b)$\int_{}^{} {{x^2}{{(3)}^{{x^3} + 1}}dx} = \int_{}^{} {3{x^2}.\,{{(3)}^{{x^3}}}dx} $
Putting ${x^3} = t \Rightarrow 3{x^2}\,dx = dt,$ we get
$ = \int_{}^{} {{3^t}dt = \frac{{{3^t}}}{{\log 3}} + c = \frac{{{{(3)}^{{x^3}}}}}{{\log 3}} + c} $.
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