Answer
$\approx 0.7594$
Work Step by Step
We are given the integral $I=\int_0^1 x\sqrt {2x+1} \ dx$
We will solve the given integral by using u-substitution method.
Let us consider that $u=2x+1 \implies dx=\dfrac{du}{2}$
So, we can write as: $\int_0^1 x\sqrt {2x+1} \ dx=\int_0^1 (\dfrac{u-1}{2})u^{1/2} \dfrac{\ du}{2}\\ =\dfrac{1}{4} \int_0^1 (u^{3/2}-u^{1/2}) \ du$
Now, use the following formula such as:
$\int x^n \ dx=\dfrac{x^{n+1}}{n+1}+C$
Therefore, we have:$ \dfrac{1}{4} \int_0^1 (u^{3/2}-u^{1/2}) \ du =\dfrac{1}{4}[\dfrac{2u^{5/2}}{5}-\dfrac{2u^{3/2}}{3}]_0^1+C$
or, $=[\dfrac{(2x+1)^{5/2}}{10}-\dfrac{(2x+1)^{3/2}}{6}]_0^1+C$
or, $=[\dfrac{(2+1)^{5/2}}{10}-\dfrac{(2+1)^{3/2}}{6}]-[\dfrac{(0+1)^{5/2}}{10}-\dfrac{(0+1)^{3/2}}{6}]$
or, $ \approx 0.7594$
