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7.2 definite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.2 definite integral1 Mark
MCQ
$\int_0^2 {\frac{{{x^3}\,dx}}{{{{({x^2} + 1)}^{\frac{3}{2}}}}}} = $
  • A
    ${(\sqrt 2 - 1)^2}$
  • B
    $\frac{{{{(\sqrt 2 - 1)}^2}}}{{\sqrt 2 }}$
  • C
    $\frac{{\sqrt 2 - 1}}{{\sqrt 2 }}$
  • ✓
    None of these
✓

Answer

Correct option: D.
None of these
d
(d) Put $t = {x^2} + 1 \Rightarrow dt = 2x\,dx$

$\int_0^2 {\frac{{{x^3}}}{{{{({x^2} + 1)}^{3/2}}}}dx = \frac{1}{2}} \int_1^5 {\frac{{(t - 1)}}{{{t^{3/2}}}}dt = \frac{1}{2}\int_1^5 {[{t^{ - 1/2}} - {t^{ - 3/2}}]\,dt} } $

$ = \frac{1}{2}\left[ {2\sqrt t + 2\frac{1}{{\sqrt t }}} \right]_1^5 $

$= \frac{1}{2}\left[ {2\sqrt 5 + \frac{2}{{\sqrt 5 }} - 2 - 2} \right]$

$ = \left[ {\sqrt 5 + \frac{1}{{\sqrt 5 }} - 2} \right] $

$= \frac{{6 - 2\sqrt 5 }}{{\sqrt 5 }}$.

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