Question

$\int \frac{x^{3}dx}{\sqrt{1+x^2}}$ equals :

• A. $\frac{{\left(1+x^2\right)}^{3/2}}{3}-\sqrt{1+x^2}+c$
• B. $x^2\sqrt{1+x^2}-\frac{1}{3}\sqrt{{\left(1+x^2\right)}^3}+c$
• C. $x^2\sqrt{1+x^2}-\frac{2}{3}\sqrt{{\left(1+x^2\right)}^3}+c$
• D. none

Answer 1

Answer: (A), (C)

$1+x^2=t^2\Rightarrow I=\int \left(t^2-1\right)dt=\frac{t^3}{3}-t=\frac{{\left(1+x^2\right)}^{3/2}}{3}-\sqrt{1+x^2}$
$=\frac{\left(1+x^2\right)\sqrt{1+x^2}}{3}-\sqrt{1+x^2}\left(1+x^2-x^2\right)=x^2\sqrt{1+x^2}-\frac{2}{3}{\left(1+x^2\right)}^{3/2}$