Question

Evaluate the following:
$\int (3x+1)\sqrt{2x-1}dx$

Answer 1

$I=\int (3x+1)\sqrt{2x-1}dxPut2x-1=t2dx=dtdx=\frac{1}{2}dt\therefore 3x+1=(t+1)\frac{3}{2}+1\therefore I=\frac{1}{2}\int (\frac{3}{2}t+\frac{3}{2}+1)\sqrt{t}dtI=\frac{1}{2}\int (\frac{3}{2}{t}^{\frac{3}{2}}+\frac{5}{2}{t}^{\frac{1}{2}})dtI=\frac{1}{2}(\frac{3}{2}\times \frac{2}{5}{t}^{\frac{5}{2}}+\frac{5}{2}\times \frac{2}{3}{t}^{\frac{3}{2}})+C(\because \int x^{n}dx=\frac{1.x^{n+1}}{n+1})I=\frac{3}{10}{t}^{\frac{5}{2}}+\frac{5}{6}{t}^{\frac{3}{2}}+CI=\frac{3}{10}{(2x-1)}^{{}^{\frac{5}{2}}}+\frac{5}{6}{(2x-1)}^{\frac{3}{2}}+C$