If ∫x³ dx/(√(1+x²) = a (1+ x²)³/² +b√(1+x²)+ C, then a and b is given by:$\\ \mathbf{\;A)\; a=\frac{1}{3} , b=1.\; B)\; a=-\frac{1}{3}, b=1\; \\ C)\; a=-\frac{1}{3}, b=-1\; D)\; a=\frac{1}{3},b=-1} $

 

We have 

$\displaystyle \int \frac{x^{3}}{\sqrt{(1+x^{2})}} \; dx= a(1+x^{2})^{\frac{3}{2}}+b\sqrt{(1+x^{2})}+C \cdots (1)$ 

In-order  to find value of a and b , at first we have to evaluate the integral $\displaystyle \int \frac{x^{3}}{\sqrt{(1+x^{2})}} \; dx$ .

Now

 $\displaystyle \int \frac{x^{3}}{\sqrt{(1+x^{2})}} \; dx=\int \frac{x^{2}\;.\: x}{\sqrt{(1+x^{2})}} \; dx$ 

Take 

$\displaystyle 1+x^{2}=u\Rightarrow 2x=\frac{\mathrm{d} u}{\mathrm{d} x}\mathbf{(Differentiating \; w.r.t \; x)}\\  \Rightarrow x \; dx=\frac{du}{2}$ 

Now by substituting $\displaystyle 1+x^{2}=u$  we will get

\[\begin{align*} \int \frac{x^{3}}{\sqrt{(1+x^{2})}} \; dx &= \int \frac{u-1}{\sqrt{u}} \frac{du}{2}\\ &= \frac{1}{2}\int \frac{u}{\sqrt{u}}- \frac{1}{\sqrt{u}} \; du\\ &=\frac{1}{2}[\int \frac{\sqrt{u}}{1}\; du-\int \frac{1}{\sqrt{u}}\; du] \\ &= \frac{1}{2}[\int u^{\frac{1}{2}}\; du-\int u^{-\frac{1}{2}}\; du]\\ &=\frac{1}{2}[\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}]+C \\ &=\frac{1}{2}[\frac{u^{3/2}}{3/2}-\frac{u^{1/2}}{1/2}] +C \\&= \frac{u^{3/2}}{3}-u^{1/2}+C \end{align*}\]

Now by providing value $\displaystyle u=1+x^{2}$  we will get:

$\displaystyle \int\frac{ x^{3}}{\sqrt{(1+x^{2})}} \; dx=\frac{1}{3}\left (1+x^{2} \right )^{\frac{3}{2}}-\left ( 1+x^{2} \right )^{\frac{1}{2}} +C \cdots(2)$ 

Comparing (1) and (2) we will get

$\displaystyle \mathbf{a=1/3\; and \; b=-1}$ 

Hence correct option :D

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