Question

Evaluate the integral

Intergral xsquared +1over square root of xcubed +3x whole thing to dx

using appropriate substitution.

Answer 1

We can evaluate this integral by substitution.

Set $u=x^{3}+3x$. This means $\frac {du}{dx}=3x^{2}+3$, or $dx=\frac{du}{3(x^2+1)}.$

Now

$\int \frac{x^2+1}{\sqrt{x^3+3x}}dx=\int \frac{x^2+1}{\sqrt{u}} \frac{du}{3(x^2+1)}= \int \frac{du}{3 \sqrt{u}}=$

$=\frac{1}{3} \int \frac{du}{\sqrt{u}}= \frac{1}{3} \cdot 2\sqrt{u}+C= \frac{2}{3}\sqrt{u}+C=$

$=\frac{2}{3}\sqrt{x^3+3x}+C.$