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### Answers (1)

Answer:

$a=\frac{1}{3} ; b=-1$

Given:

$\int \frac{x^{3}}{\sqrt{1+x^{2}}} d x=a\left(1+x^{2}\right)^{\frac{3}{2}}+b \sqrt{1+x^{2}}+C$

Hint:

Using $\int x^{n} d x$.

Explanation:

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

\begin{aligned} &=\int \frac{x^{2} x}{\sqrt{1+x^{2}}} d x \\ &=\int \frac{(t-1)}{\sqrt{t}} \cdot \frac{d t}{2} \end{aligned}                                                                        $\text { [ Put } \left.1+x^{2}=t \Rightarrow x^{2}=t-1 \Rightarrow 2 x d x=d t\right]$

\begin{aligned} &=\frac{1}{2} \int \frac{t-1}{\sqrt{t}} d t \\ &=\frac{1}{2} \int \sqrt{t} d t-\frac{1}{2} \int \frac{1}{\sqrt{t}} d t \\ &=\frac{1}{2} \int(t)^{\frac{1}{2}} d t-\frac{1}{2} \int(t)^{\frac{-1}{2}} d t \\ &=\frac{1}{2}\left[\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]-\frac{1}{2}\left[\frac{(t)^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}\right]+C \\ &=\frac{1}{2} \times \frac{t^{\frac{3}{2}}}{\frac{3}{2}}-\frac{1}{2} \times \frac{(t)^{\frac{1}{2}}}{\frac{1}{2}}+C \\ &=\frac{1}{3} t^{\frac{3}{2}}-t^{\frac{1}{2}}+C \end{aligned}

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

Acc. to given  $I=a\left(1+x^{2}\right)^{\frac{8}{2}}+b \sqrt{1+x^{2}}+C$

$\Rightarrow \frac{1}{3}\left(1+x^{2}\right)^{\frac{3}{2}}-\sqrt{1+x^{2}}+C=a\left(1+x^{2}\right)^{\frac{3}{2}}+b \sqrt{1+x^{2}}+C$

On comparing, we get

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

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