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Definite Integration — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integration2 Marks
MCQ
$\int_0^2 \frac{x^3 d x}{\left(x^2+1\right)^{\frac{3}{2}}}=$
  • A
    $(\sqrt{2}-1)^2$
  • B
    $\frac{(\sqrt{2}-1)^2}{\sqrt{2}}$
  • C
    $\frac{\sqrt{2}-1}{\sqrt{2}}$
  • ✓
    $\frac{6-2 \sqrt{5}}{\sqrt{5}}$
✓

Answer

Correct option: D.
$\frac{6-2 \sqrt{5}}{\sqrt{5}}$
(D)
Put $x^2+1= t \Rightarrow 2 x d x= dt$
When $x=0, t =1$ and when $x=2, t =5$
$\therefore \int_0^2 \frac{x^3}{\left(x^2+1\right)^{\frac{3}{2}}} d x=\frac{1}{2} \int_1^5 \frac{( t -1)}{ t ^{\frac{3}{2}}} dt$
$=\frac{1}{2} \int_1^5\left( t ^{\frac{-1}{2}}- t ^{\frac{-3}{2}}\right) dt \\ =\frac{1}{2}\left[2 \sqrt{ t }+2 \frac{1}{\sqrt{ t }}\right]_1^5 \\ =\frac{1}{2}\left[2 \sqrt{5}+\frac{2}{\sqrt{5}}-2-2\right] \\ =\sqrt{5}+\frac{1}{\sqrt{5}}-2 \\ =\frac{6-2 \sqrt{5}}{\sqrt{5}}$

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