Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \left(5x+3\right)\sqrt{2x-1}dx \\ \mathrm{Putting}2x-1=t \\ \Rightarrow 2x=t+1 \\ \Rightarrow x=\frac{t+1}{2} \end{gathered}$$

$$\begin{gathered} \&2dx=dt \\ \Rightarrow dx=\frac{dt}{2} \end{gathered}$$

$$\begin{gathered} \therefore I=\int \left[5\left(\frac{t+1}{2}\right)+3\right]·\sqrt{t}·\frac{dt}{2} \\ =\int \left(\frac{5t}{2}+\frac{5}{2}+3\right)\times \frac{\sqrt{t}dt}{2} \\ =\frac{1}{4}\int \left(5t+11\right)t^{\frac{1}{2}}dt \\ =\frac{1}{4}\int \left(5t^{\frac{3}{2}}+11t^{\frac{1}{2}}\right)dt \\ =\frac{1}{4}\left[5\frac{t^{{\displaystyle \frac{3}{2}}+1}}{{\displaystyle \frac{3}{2}}+1}+11\frac{t^{{\displaystyle \frac{1}{2}}+1}}{{\displaystyle \frac{1}{2}}+1}\right]+\mathrm{C} \\ =\frac{1}{4}\times \frac{2}{5}\times 5t^{\frac{5}{2}}+\frac{1}{4}\times 11\times \frac{2}{3}{t}^{\frac{3}{2}}+\mathrm{C} \\ =\frac{1}{2}{t}^{\frac{5}{2}}+\frac{11}{6}{t}^{\frac{3}{2}}+\mathrm{C} \\ =\frac{t^{{\displaystyle \frac{3}{2}}}}{2}\left[t+\frac{11}{3}\right]+\mathrm{C} \\ =\frac{t^{{\displaystyle \frac{3}{2}}}}{2}\left[\frac{3t+11}{3}\right]+\mathrm{C} \\ =\frac{{\left(2x-1\right)}^{{\displaystyle \frac{3}{2}}}}{2}\left[\frac{3\left(2x-1\right)+11}{3}\right]+\mathrm{C}\left[\because t=2x-1\right] \\ =\frac{{\left(2x-1\right)}^{{\displaystyle \frac{3}{2}}}}{2}\left[\frac{6x-3+11}{3}\right]+\mathrm{C} \\ ={\left(\frac{2x-1}{2}\right)}^{\frac{3}{2}}\left[\frac{2\left(3x+4\right)}{3}\right]+\mathrm{C} \\ =\frac{{\left(2x-1\right)}^{{\displaystyle \frac{3}{2}}}\left(3x+4\right)}{3}+\mathrm{C} \end{gathered}$$