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