MCQ
$\int_0^\infty {\frac{{{x^3}\,dx}}{{{{({x^2} + 4)}^2}}} = } $
- A$0$
- ✓$\infty $
- C$\frac{1}{2}$
- DNone of these
✓
Answer
Correct option: B.
$\infty $
b
(b) $\int_0^\infty {\frac{{{x^3}dx}}{{{{({x^2} + 4)}^2}}} = \frac{1}{2}} \int_0^\infty {\frac{{{x^2}2x\,dx}}{{{{({x^2} + 4)}^2}}}dx} $
(b) $\int_0^\infty {\frac{{{x^3}dx}}{{{{({x^2} + 4)}^2}}} = \frac{1}{2}} \int_0^\infty {\frac{{{x^2}2x\,dx}}{{{{({x^2} + 4)}^2}}}dx} $
$ = \frac{1}{2}\int_0^\infty {\frac{t}{{{{(t + 4)}^2}}}dt} $,
[Putting ${x^2} = t$]
$ = \frac{1}{2}\int_0^\infty {\left[ {\frac{1}{{t + 4}} - \frac{4}{{{{(t + 4)}^2}}}} \right]dt = \frac{1}{2}\left[ {\log (t + 4) + \frac{4}{{t + 4}}} \right]_0^\infty } $
$ = \frac{1}{2}\left[ {\log \infty + 0 - (\log 4 + 1)} \right] = \infty $.
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