## University Calculus: Early Transcendentals (3rd Edition)

$$\int x\sqrt{2x-3}dx=\frac{(2x-3)^{3/2}(x+1)}{5}+C$$
$$A=\int x\sqrt{2x-3}dx=\int x(2x-3)^{1/2}dx$$ Using Formula 22, which states that $$\int x(ax+b)^ndx=\frac{(ax+b)^{n+1}}{a^2}\Big[\frac{ax+b}{n+2}-\frac{b}{n+1}\Big]+C$$ with $a=2$, $b=-3$ and $n=1/2$ here, we have $$A=\frac{(2x-3)^{3/2}}{2^2}\Big[\frac{2x-3}{\frac{5}{2}}-\frac{(-3)}{\frac{3}{2}}\Big]+C$$ $$A=\frac{(2x-3)^{3/2}}{4}\Big[\frac{4x-6}{5}+2\Big]+C$$ $$A=\frac{(2x-3)^{3/2}}{4}\Big[\frac{4x+4}{5}\Big]+C$$ $$A=\frac{(2x-3)^{3/2}(x+1)}{5}+C$$