Evaluate the following integrals:
Question:

Evaluate $\int x \sqrt{2 x+3} d x$

Solution:

In this question we write $x \sqrt{2 x+3}$ as

$x \sqrt{2 x+3}=\frac{2 x \sqrt{2 x+3}}{2}$

$=\frac{(2 x+3-3) \sqrt{2 x+3}}{2}$

$=\frac{(2 x+3) \sqrt{2 x+3}-3 \sqrt{2 x+3}}{2}$

$=\frac{(2 x+3)^{\frac{3}{2}}-3 \sqrt{2 x+3}}{2}$

$y=\int x \sqrt{2 x+3} d x$

$y=\int \frac{(2 x+3)^{\frac{3}{2}}-3 \sqrt{2 x+3}}{2} d x$

Using formula $\int(a x+b)^{n} d x=\frac{(a x+b)^{n+1}}{a(n+1)}$

$y=\frac{(2 x+3)^{\frac{5}{2}}}{2 \times 2 \times \frac{5}{2}}-\frac{3(2 x+3)^{\frac{3}{2}}}{2 \times 2 \times \frac{3}{2}}+c$

$y=\frac{(2 x+3)^{\frac{5}{2}}}{10}-\frac{(2 x+3)^{\frac{3}{2}}}{2}+c$