Question

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

Answer 1

\[\text{Let I }= \int\left( \frac{3x + 5}{\sqrt{7x + 9}} \right)dx\]
\[Putting\ 7x + 9 = t\]
\[ \Rightarrow x = \frac{t - 9}{7}\]

\[\text{and}\ 7dx = dt\]
\[ \Rightarrow dx = \frac{dt}{7}\]

\[\therefore I = \int\left( \frac{3\left( \frac{t - 9}{7} \right) + 5}{\sqrt{t}} \right)dt\]
\[ = \int\left( \frac{3}{7}\frac{t}{\sqrt{t}} - \frac{27}{7\sqrt{t}} + \frac{5}{\sqrt{t}} \right)\frac{dt}{7}\]

` = 3/(7× 7) ∫ t^{1/2} dt - 27/{7× 7 }   ∫  t ^{-1/2} dt + 5/7 ∫   t ^{-1/2}dt `
\[ = \frac{3}{7 \times 7}\left[ \frac{t^\frac{1}{2} + 1}{\frac{1}{2} + 1} \right] - \frac{27}{7 \times 7}\left[ \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} \right] + \frac{5}{7}\left[ \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} \right] + C\]

\[ = \frac{2}{7 \times 7} t^\frac{3}{2} - \frac{27}{7 \times 7} \times \text{2 }   t^\frac{1}{2} + \frac{10\sqrt{t}}{7} + C\]
\[ = \frac{2}{7 \times 7} \left( 7x + 9 \right)^\frac{3}{2} - \frac{54}{7 \times 7} \left( 7x + 9 \right)^\frac{1}{2} + \frac{10}{7}\sqrt{7x + 9} + C \left[ \because t = 7x + 9 \right]\]
\[ = \frac{2}{7 \times 7} \left( 7x + 9 \right)^\frac{3}{2} + \left( 10 - \frac{54}{7} \right) \frac{\sqrt{7x + 9}}{7} + C\]
\[ = \frac{2}{7} \left( 7x + 9 \right)^\frac{3}{2} + \left( \frac{70 - 54}{7} \right) \frac{\sqrt{7x + 9}}{7} + C\]
\[ = \frac{2}{7 \times 7} \left( 7x + 9 \right)^\frac{3}{2} + \frac{16}{7 \times 7}\sqrt{7x + 9} + C\]
\[ = \frac{2}{7 \times 7}\left[ \left( 7x + 9 \right)^\frac{1}{2} \left[ 7x + 9 + 8 \right] \right] + C\]
\[ = \frac{2}{49}\left[ \left( 7x + 9 \right)^\frac{1}{2} \left( 7x + 17 \right) \right] + C\]