Question

$\int \frac{x^5}{\sqrt{1+x^3}}dx=$

• A. $\frac{2}{9}(1+x^3{)}^{3/2}+c$
• B. $\frac{2}{9}(1+x^3{)}^{3/2}+c+\frac{2}{3}(1+x^3{)}^{1/2}+c$
• C. $\frac{2}{9}(1+x^3{)}^{3/2}+c-\frac{2}{3}(1+x^3{)}^{1/2}+c$
• D. None of these

Answer 1

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

Put $1+x^3=t^2\Rightarrow 3x^{2}dx=2tdt$ and $x^3=t^2-1$
So, $\int \frac{x^5}{\sqrt{1+x^3}}dx=\int \frac{x^2.x^3}{\sqrt{1+x^3}}dx$
$=\frac{2}{3}\int \frac{(t^2-1).tdt}{t}=\frac{2}{3}\int (t^2-1)dt=\frac{2}{3}[\frac{t^3}{3}-t]+c$
$=\frac{2}{3}[\frac{(1+x^3{)}^{3/2}}{3}-(1+x^3{)}^{1/2}]+c$