Answer
$$\int_0^4\frac{x}{\sqrt{1+2x}}dx=\frac{10}{3}$$
Work Step by Step
To evaluate the integral $$\int_0^4\frac{x}{\sqrt{1+2x}}dx$$
we will use substitution $1+2x=t$ which gives us $2dx=dt\Rightarrow dx=\frac{dt}{2}$ and $x=\frac{t-1}{2}$. New integration bounds would be: for $x=0$ we have $t=1$ and for $x=4$ we have $t=9.$ Putting this into the integral we get:
$$\int_0^4\frac{x}{\sqrt{1+2x}}dx=\int_1^9\frac{1}{\sqrt t}\frac{t-1}{2}\frac{dt}{2}=\frac{1}{4}\int_1^9\frac{t}{\sqrt t}dt-\frac{1}{4}\int_1^9\frac{1}{\sqrt t}dt=
\frac{1}{4}\int_1^9 t^{1/2}dt-\frac{1}{4}\int_1^9 t^{-1/2}dt=
\left.\frac{1}{4}\frac{t^{3/2}}{\frac{3}{2}}\right|_1^9-\left.\frac{1}{4}\frac{t^{1/2}}{\frac{1}{2}}\right|_1^9=
\frac{1}{4}\frac{2}{3}(9^{3/2}-1^{3/2})-\frac{1}{4}\cdot2(9^{1/2}-1^{1/2})=
\frac{1}{6}(27-1)-\frac{1}{2}(3-1)=
\frac{13}{3}-1=\frac{10}{3}$$
