First slide

Methods of integration

Question

$\int \frac{x^{3} d x}{\sqrt{1 + x^{2}}} is equal to$

Moderate

A

$\frac{1}{3} \sqrt{1 + x^{2}} (2 + x^{2}) + C$
B

$\frac{1}{3} \sqrt{1 + x^{2}} (x^{2} - 1) + C$
C

$\frac{1}{3} {(1 + x^{2})}^{3 / 2} + C$
D

$\frac{1}{3} \sqrt{1 + x^{2}} (x^{2} - 2) + C$

Solution

$I = \int \frac{x^{3} d x}{\sqrt{1 + x^{2}}} = \int \frac{x \cdot x^{2} d x}{\sqrt{1 + x^{2}}}$
$Let t = \sqrt{1 + x^{2}} \Rightarrow \frac{d t}{d x} = \frac{x}{\sqrt{1 + x^{2}}}$
$Now, I = \int (t^{2} - 1) d t = \frac{t^{3}}{3} - t + C = \frac{t}{3} (t^{2} - 3) + C$
$∴ I = \frac{1}{3} \sqrt{1 + x^{2}} (x^{2} - 2) + C$

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