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  4. If ∫(x3 dx/√1+x2)= a(1+x2) 3/2 +b √1+x2+ C, then

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Q. If $\int\frac{x^{3} dx}{\sqrt{1+x^{2}}}= a\left(1+x^{2}\right)^{ {3}/{2}} +b \sqrt{1+x^{2}}+ C$, then

Integrals

Solution:

We have, $I = \int\frac{ x^{3} dx}{\sqrt{1+x^{2} }}$

Put $x^{2} = t \Rightarrow 2x \, dx = dt$

$ \Rightarrow I= \frac{1}{2} \int \frac{t\, dt}{\sqrt{1+t}} = \frac{1}{2} \int \frac{t+ 1-1}{\sqrt{1+t}} dt $

$= \frac{1}{2} \int \left(\sqrt{1+t} -\left(1+t\right)^{-{1}/{2}}\right) dt$

$ = \frac{1}{2}\times \frac{2}{3} \left(1+t\right)^{{3}/{2} } -\frac{1}{2} \times 2\left(1+t\right)^{{1}/{2}} + C $

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

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