Answer
$\dfrac{(2x-3)^{3/2}(x+1)}{5}+C$
Work Step by Step
By using Formula 22, we get:
$\int x(ax+b)^ndx=\dfrac{(ax+b)^{n+1}}{a^2} \space [\dfrac{ax+b}{n+2}-\dfrac{b}{n+1}]+ \space C$
Let $I=\dfrac{(2x-3)^{3/2}}{2^2}[\dfrac{2x-3}{\dfrac{5}{2}}-\dfrac{(-3)}{\dfrac{3}{2}}]+C \\=\dfrac{(2x-3)^{3/2}}{4} [\dfrac{4x-6}{5}+2]+C\\=\dfrac{(2x-3)^{3/2}(x+1)}{5}+C$
