If x 3 1 + x 2 ? d x = a ( 1 + x 2 ) 3 / 2 + b 1 + x 2 + C , then:

Question

If $\int \frac{x^{3}}{\sqrt{1 + x^{2}}} d x = a {(1 + x^{2})}^{3 / 2} + b \sqrt{1 + x^{2}} + C,$ then:

Smart Achievers · Accepted Answer

(4)

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

Explanation:

Let $I = \int \frac{x^{3}}{\sqrt{1 + x^{2}}} d x$

$= a {(1 + x^{2})}^{3 / 2} + b \sqrt{1 + x^{2}} + C$

$∵$ $I = \int \frac{x^{3}}{\sqrt{1 + x^{2}}} d x$

$= \int \frac{x^{2} \cdot x}{\sqrt{1 + x^{2}}} d x$

Put $1 + x^{2} = t^{2}$

$\Rightarrow$ $2 x d x = 2 t d t$

$I = \int \frac{t (t^{2} - 1)}{t} d t$

$= \int (t^{2} - 1) d t = \frac{t^{3}}{3} - t + C$

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

By comparison, $a = \frac{1}{3}$ and $b = - 1$

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