Question

# $\int \left(5x+3\right)\sqrt{2x-1}dx$

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Solution

## $\mathrm{Let}I=\int \left(5x+3\right)\sqrt{2x-1}dx\phantom{\rule{0ex}{0ex}}\mathrm{Putting}2x-1=t\phantom{\rule{0ex}{0ex}}⇒2x=t+1\phantom{\rule{0ex}{0ex}}⇒x=\frac{t+1}{2}$ $&2dx=dt\phantom{\rule{0ex}{0ex}}⇒dx=\frac{dt}{2}$ $\therefore I=\int \left[5\left(\frac{t+1}{2}\right)+3\right]·\sqrt{t}·\frac{dt}{2}\phantom{\rule{0ex}{0ex}}=\int \left(\frac{5t}{2}+\frac{5}{2}+3\right)×\frac{\sqrt{t}dt}{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\int \left(5t+11\right){t}^{\frac{1}{2}}dt\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\int \left(5{t}^{\frac{3}{2}}+11{t}^{\frac{1}{2}}\right)dt\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\left[5\frac{{t}^{\frac{3}{2}+1}}{\frac{3}{2}+1}+11\frac{{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]+\mathrm{C}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}×\frac{2}{5}×5{t}^{\frac{5}{2}}+\frac{1}{4}×11×\frac{2}{3}{t}^{\frac{3}{2}}+\mathrm{C}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}{t}^{\frac{5}{2}}+\frac{11}{6}{t}^{\frac{3}{2}}+\mathrm{C}\phantom{\rule{0ex}{0ex}}=\frac{{t}^{\frac{3}{2}}}{2}\left[t+\frac{11}{3}\right]+\mathrm{C}\phantom{\rule{0ex}{0ex}}=\frac{{t}^{\frac{3}{2}}}{2}\left[\frac{3t+11}{3}\right]+\mathrm{C}\phantom{\rule{0ex}{0ex}}=\frac{{\left(2x-1\right)}^{\frac{3}{2}}}{2}\left[\frac{3\left(2x-1\right)+11}{3}\right]+\mathrm{C}\left[\because t=2x-1\right]\phantom{\rule{0ex}{0ex}}=\frac{{\left(2x-1\right)}^{\frac{3}{2}}}{2}\left[\frac{6x-3+11}{3}\right]+\mathrm{C}\phantom{\rule{0ex}{0ex}}={\left(\frac{2x-1}{2}\right)}^{\frac{3}{2}}\left[\frac{2\left(3x+4\right)}{3}\right]+\mathrm{C}\phantom{\rule{0ex}{0ex}}=\frac{{\left(2x-1\right)}^{\frac{3}{2}}\left(3x+4\right)}{3}+\mathrm{C}$

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