Question

$\int \frac{8x+13}{\sqrt{4x+7}}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \left(\frac{8x+13}{\sqrt{4x+7}}\right)dx \\ \mathrm{Putting}4x+7=t \\ \Rightarrow x=\frac{t-7}{4} \\ \Rightarrow 4dx=dt \\ \Rightarrow dx=\frac{dt}{4} \\ \therefore I=\frac{1}{4}\int \left\{\frac{8\left({\displaystyle \frac{t-7}{4}}\right)+13}{\sqrt{t}}\right\}dt \\ =\frac{1}{4}\int \left(\frac{2t-14+13}{\sqrt{t}}\right)dt \\ =\frac{1}{4}\int \left(\frac{2t-1}{\sqrt{t}}\right)dt \\ =\frac{1}{4}\int \frac{2t}{\sqrt{t}}dt-\frac{1}{4}\int \frac{dt}{\sqrt{t}} \\ =\frac{1}{2}\int t^{\frac{1}{2}}dt-\frac{1}{4}\int t^{-\frac{1}{2}}dt \\ =\frac{1}{2}\left[\frac{t^{{\displaystyle \frac{1}{2}}+1}}{{\displaystyle \frac{1}{2}}+1}\right]-\frac{1}{4}\left[\frac{t^{-{\displaystyle \frac{1}{2}}+1}}{-{\displaystyle \frac{1}{2}}+1}\right]+C \\ =\frac{1}{2}\times \frac{2}{3}{t}^{\frac{3}{2}}-\frac{2}{4}{t}^{\frac{1}{2}}+C \\ =\frac{1}{3}{t}^{\frac{3}{2}}-\frac{2}{4}{t}^{\frac{1}{2}}+C \\ =\frac{1}{3}{\left(4x+7\right)}^{\frac{3}{2}}-\frac{1}{2}{\left(4x+7\right)}^{\frac{1}{2}}+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\mathit{\because }t=4x+7\right) \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\frac{1}{3}\mathit{}{\left(4x+7\right)}^{\frac{\mathit{3}}{\mathit{2}}}\mathit{-}\sqrt{4x+7}+C \end{gathered}$$