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

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