Question

Evaluate the following integrals:

${\int }_2^4\frac{x^2+x}{\sqrt{2x+1}}dx$

Answer 1

Let I = ${\int }_2^4\frac{x^2+x}{\sqrt{2x+1}}dx$

Put 2x + 1 = z<sup>2

$$\begin{gathered} \Rightarrow 2dx=2zdz \\ \Rightarrow dx=zdz \end{gathered}$$</sup>

When $x\to 2,z\to \sqrt{5}$

When $x\to 4,z\to 3$

$$\begin{gathered} \therefore I={\int }_{\sqrt{5}}^3\frac{{\left({\displaystyle \frac{z^2-1}{2}}\right)}^2+{\displaystyle \frac{z^2-1}{2}}}{z}\times zdz \\ \Rightarrow I={\int }_{\sqrt{5}}^3\frac{z^4-2z^2+1+2z^2-2}{4}dz \\ \Rightarrow I=\frac{1}{4}{\int }_{\sqrt{5}}^3\left(z^4-1\right)dz \\ \Rightarrow I=\frac{1}{4}\times {\overline{)\left(\frac{z^5}{5}-z\right)}}_{\sqrt{5}}^3 \end{gathered}$$
$$\begin{gathered} \Rightarrow I=\frac{1}{4}\left[\left(\frac{243}{5}-3\right)-\left(\frac{25\sqrt{5}}{5}-\sqrt{5}\right)\right] \\ \Rightarrow I=\frac{1}{4}\times \frac{228}{5}-\frac{1}{4}\times 4\sqrt{5} \\ \Rightarrow I=\frac{57}{5}-\sqrt{5} \end{gathered}$$