More definite integrals using u substitution

$\begin{aligned} I & = \int_0^4 \frac x{\sqrt{1+2x}} dx & \small \color{#3D99F6} \text{Let }u^2 = 1+2x \implies x = \frac {u^2-1}2 \\ & = \int_1^3 \frac {u(u^2-1)}{2u} du & \small \color{#3D99F6} \implies 2u\ du = 2dx \\ & = \frac 12 \int_1^3 (u^2 - 1)\ du \\ & = \frac 12 \left[\frac {u^3}3 - u \right]_1^3 \\ & = \frac 12 \left[6 + \frac 23\right] \\ & = \frac {10}3 \approx \boxed{3.333} \end{aligned}$

Let $u$ be $1+2x$ , therefore $du=2 dx$ .

Therefore $x=\frac{u-1}{2}$ .

Substituting these values into the integral, the values inside the integral are:

$\Large \frac{u-1}{2 \sqrt{u}} \frac{1}{2}du$

Taking $\frac{1}{4}$ outside the integral, the integral becomes:

${\displaystyle \frac{1}{4} \int_{1}^{9} \frac{u-1}{\sqrt{u}} du }$

The bottom limit changes to 1 because when $x=0$ , $u=1+2x=1$

The top limit changes to 9 because when $x=4$ , $u=1+2x=9$

The integral in terms of $u$ can be simplified to:

${\displaystyle \frac{1}{4} \int_{1}^{9} (\frac{-1}{\sqrt{u}}+ \sqrt{u} ) du }$

Evaluating the simplified integral, the answer becomes $\frac{10}{3}$