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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS1 Mark
MCQ
Choose the correct option from given four options:
If $\frac{\text{x}^3\text{dx}}{\sqrt{1+\text{x}^2}}=\text{a}(1+\text{x}^2)^{\frac{3}{2}}+\text{b}\sqrt{1+\text{x}^2}+\text{C},$ then:
  • A
    $\text{a}=\frac{1}{3},\text{b}=1$
  • B
    $\text{a}=\frac{-1}{3},\text{b}=1$
  • C
    $\text{a}=\frac{-1}{3},\text{b}=-1$
  • ✓
    $\text{a}=\frac{1}{3},\text{b}=-1$
✓

Answer

Correct option: D.
$\text{a}=\frac{1}{3},\text{b}=-1$
Given $\text{I}=\int\frac{\text{x}^3}{\sqrt{1+\text{x}^2}}\text{dx}=\text{a}(1+\text{x}^2)^{\frac{3}{2}}+\text{b}\sqrt{1+\text{x}^2}+\text{C}$
$\text{I}=\int\frac{\text{x}^3}{\sqrt{1+\text{x}^2}}\text{dx}$
Put $1+\text{x}^2=\text{t}^2$
$\Rightarrow\ 2\text{x dx}=2\text{t dt}$
$\therefore\ \text{I}\int\frac{\text{t}(\text{t}^2-1)}{\text{t}}\text{dt}$ $=\frac{\text{t}^3}{3}-\text{t}+\text{C}=\frac{1}{3}(1+\text{x}^2)^{\frac{3}{2}}-\sqrt{1+\text{x}^2}+\text{C}$
$\therefore\ \text{a}=\frac{1}{3}$ and $\text{b}=-1$

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