Question

$\int \frac{x^3}{\sqrt{1+x^2}}\mathrm{d}x=a{\left(1+x^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+b\sqrt{1+x^2}+C$, then

$$\begin{gathered} \left(\mathrm{a}\right)a=\frac{1}{3},b=1 \\ \left(\mathrm{b}\right)a=-\frac{1}{3},b=1 \\ \left(\mathrm{c}\right)a=-\frac{1}{3},b=-1 \\ \left(\mathrm{d}\right)a=\frac{1}{3},b=-1 \end{gathered}$$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{x^3}{\sqrt{1+x^2}}\mathrm{d}x \\ =\int \frac{x.x^2}{\sqrt{1+x^2}}\mathrm{d}x \\ \mathrm{Let}\left(1+x^2\right)=t \\ \mathrm{On}\mathrm{differentiating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ 2x\mathrm{d}x=\mathrm{d}t \\ \therefore I=\frac{1}{2}\int \frac{t-1}{\sqrt{t}}\mathrm{d}t \\ =\frac{1}{2}\int \left(\frac{t}{\sqrt{t}}-\frac{1}{\sqrt{t}}\right)\mathrm{d}t \\ =\frac{1}{2}\int \left(t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-t^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)\mathrm{d}t \\ =\frac{1}{2}\left(\frac{2}{3}{t}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{2}{1}{t}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)+C \\ =\left(\frac{1}{3}{\left(1+x^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\sqrt{1+x^2}\right)+C \\ \mathrm{Since},\int \frac{x^3}{\sqrt{1+x^2}}\mathrm{d}x=a{\left(1+x^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+b\sqrt{1+x^2}+C \\ \mathrm{Therefore},a=\frac{1}{3},b=-1. \\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\left(\mathrm{d}\right). \end{gathered}$$