## Thomas' Calculus 13th Edition

$\dfrac{x+3}{\sqrt{2x+3}}+C$
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 $I=\int\dfrac{x \space dx}{(2x+3)^{3/2}}=\int x(2x+3)^{-3/2}dx =\dfrac{(2x+3)^{-1/2}}{2^2} [\dfrac{2x+3}{\dfrac{1}{2}}-\dfrac{3}{-\dfrac{1}{2}}]+C \\=\dfrac{1}{4\sqrt{2x+3}} [4x+6+6 ]+C \\=\dfrac{x+3}{\sqrt{2x+3}}+C$