x^3 1+x^2 dx=a(1+x^2)^ 3 2 +b 1+x^2 +C, then: - Vidyadip

MCQ

$\int\frac{\text{x}^3}{\sqrt{1+\text{x}^2}}\text{ dx}=\text{a}(1+\text{x}^2)^{\frac{3}{2}}+\text{b}\sqrt{1+\text{x}^2}+\text{C},$ 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.

$\text{a}=\frac{1}{3},\text{ b}=-1$

$\text{I}=\int\frac{\text{x}^3}{\sqrt{1+\text{x}^2}}\text{ dx}$
$1+\text{x}^2=\text{t}$
$2\text{xdx}=\text{dt}$
$\text{x dx}=\frac{\text{dt}}{2}$
$\text{I}=\int\frac{\text{x}^2}{\sqrt{1+\text{x}^2}}\text{x dx}$
$\text{I}=\int\frac{\text{t}-1}{\sqrt{\text{t}}}\frac{\text{dt}}{2}$
$\text{I}=\frac{1}{2}\Big(\frac{2}{3}\text{t}^{\frac{3}{2}}-2\sqrt{\text{t}}\Big)+\text{C}$
$\text{I}=\frac{1}{3}(1+\text{x}^{2})^{\frac{3}{2}}-\sqrt{1+\text{x}^2}+\text{C}$
$\text{a}=\frac{1}{3},\text{ b}=-1$

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