If $\int \frac{x^3 \mathrm{~d} x}{\sqrt{1+x^2}}=\mathrm{a}\left(1+x^2\right) \sqrt{1+x^2}+\mathrm{b} \sqrt{1+x^2}+\mathrm{c}$ (where $\mathrm{c}$ is a constant of integration), then the value of $3 \mathrm{ab}$ is

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Put } 1+x^2=\mathrm{t}^2 \ & \Rightarrow 2 x \mathrm{~d} x=2 \mathrm{t} \mathrm{dt} \ & \Rightarrow x \mathrm{~d} x=\mathrm{t} d \mathrm{t} \ & \quad \int \frac{x^3 \mathrm{~d} x}{\sqrt{1+x^2}} \ & =\int \frac{\mathrm{t}^2-1}{\mathrm{t}} \cdot \mathrm{t} d \mathrm{t} \ & =\int\left(\mathrm{t}^2-1ight) \mathrm{dt} \ & =\frac{\mathrm{t}^3}{3}-\mathrm{t}+\mathrm{c} \ & =\frac{\left(1+x^2ight)^{\frac{3}{2}}}{3}-\sqrt{1+x^2}+\mathrm{c} \ & =\frac{1}{3}\left(1+x^2ight) \sqrt{1+x^2}-\sqrt{1+x^2}+\mathrm{c}\end{aligned}$ $\begin{aligned} 	herefore \quad & a=\frac{1}{3}, b=-1 \ & \Rightarrow 3 a b=-1\end{aligned}$

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

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