Question

$\int \frac{x^5}{\sqrt{1+x^3}}dx$ is equal to

• A. $\frac{2}{9}\sqrt{1+x^2}(x^3-9)+C$
• B. $\frac{2}{9}\sqrt{x^3-9}(1+x^2)+C$
• C. $\frac{2}{9}\sqrt{1+x^3}+C$
• D. $\frac{2}{9}\sqrt{1+x^3}(x^3-2)+C$
• E. $\frac{2}{9}\sqrt{1+x^2}(x^3+9)+C$

Answer 1

The correct option is A $\frac{2}{9}\sqrt{1+x^2}(x^3-9)+C$

Let $I=\int \frac{x^5}{\sqrt{1+x^3}}dx=\int \frac{x^3\cdot x^2}{\sqrt{1+x^3}}dx$
Put $1+x^3=t^2$
$\Rightarrow 3x^{2}dx=2tdt$
$\Rightarrow x^{2}dx=\frac{2}{3}tdt$
Also, $x^3=t^2-1$
$\therefore I=\frac{2}{3}\int \frac{(t^2-1)}{t}tdt$
$=\frac{2}{3}\int (t^2-1)dt$
$=\frac{2}{3}[\frac{t^3}{3}-t]+C$
$=\frac{2}{3}t[\frac{t^2}{3}-1]+C$
$\Rightarrow I=\frac{2}{9}t(t^2-3)+C$
Put $t=\sqrt{1+x^3}$
$\Rightarrow I=\frac{2}{9}\sqrt{1+x^3}(x^3-2)+C$