Question

$\int \frac{3x+5}{\sqrt{7x+9}}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \left(\frac{3x+5}{\sqrt{7x+9}}\right)dx \\ \mathrm{Putting}7x+9=t \\ \Rightarrow x=\frac{t-9}{7} \end{gathered}$$

$$\begin{gathered} \&7dx=dt \\ \Rightarrow dx=\frac{dt}{7} \end{gathered}$$

$$\begin{gathered} \therefore I=\int \left(\frac{3\left({\displaystyle \frac{t-9}{7}}\right)+5}{\sqrt{t}}\right)dt \\ =\int \left(\frac{3}{7}\frac{t}{\sqrt{t}}-\frac{27}{7\sqrt{t}}+\frac{5}{\sqrt{t}}\right)\frac{dt}{7} \\ =\frac{3}{7\times 7}\int t^{\frac{1}{2}}dt-\frac{27}{7\times 7}\int t^{-\frac{1}{2}}dt+\frac{5}{7}\int t^{-\frac{1}{2}}dt \\ =\frac{3}{7\times 7}\left[\frac{t^{{\displaystyle \frac{1}{2}}+1}}{{\displaystyle \frac{1}{2}}+1}\right]-\frac{27}{7\times 7}\left[\frac{t^{-{\displaystyle \frac{1}{2}}+1}}{-{\displaystyle \frac{1}{2}}+1}\right]+\frac{5}{7}\left[\frac{t^{-{\displaystyle \frac{1}{2}}+1}}{-{\displaystyle \frac{1}{2}}+1}\right]+\mathrm{C} \\ =\frac{2}{7\times 7}{t}^{\frac{3}{2}}-\frac{27}{7\times 7}\times 2t^{\frac{1}{2}}+\frac{10\sqrt{t}}{7}+\mathrm{C} \\ =\frac{2}{7\times 7}{\left(7x+9\right)}^{\frac{3}{2}}-\frac{54}{7\times 7}{\left(7x+9\right)}^{\frac{1}{2}}+\frac{10}{7}\sqrt{7x+9}+\mathrm{C}\left[\because t=7x+9\right] \\ =\frac{2}{7\times 7}{\left(7x+9\right)}^{\frac{3}{2}}+\left(10-\frac{54}{7}\right)\frac{\sqrt{7x+9}}{7}+\mathrm{C} \\ =\frac{2}{7}{\left(7x+9\right)}^{\frac{3}{2}}+\left(\frac{70-54}{7}\right)\frac{\sqrt{7x+9}}{7}+\mathrm{C} \\ =\frac{2}{7\times 7}{\left(7x+9\right)}^{\frac{3}{2}}+\frac{16}{7\times 7}\sqrt{7x+9}+\mathrm{C} \\ =\frac{2}{7\times 7}\left[{\left(7x+9\right)}^{\frac{1}{2}}\left[7x+9+8\right]\right]+\mathrm{C} \\ =\frac{2}{49}\left[{\left(7x+9\right)}^{\frac{1}{2}}\left(7x+17\right)\right]+\mathrm{C} \end{gathered}$$