Question

\[\int\frac{2 - 3x}{\sqrt{1 + 3x}} dx\]

Answer 1

\[ \text{Let I} = \int\left( \frac{2 - 3x}{\sqrt{1 + 3x}} \right)dx\]

Putting 1 + 3x = t
⇒ 3x = t – 1

\[\text{and}\ 3dx = dt\]
\[ \Rightarrow dx = \frac{dt}{3}\]

\[\therefore I = \int\left( \frac{2 - \left( t - 1 \right)}{\sqrt{t}} \right)dt\]
\[ = \int\left( \frac{3 - t}{\sqrt{t}} \right)dt\]
\[ = \int\left( 3 t^{- \frac{1}{2}} - t^\frac{1}{2} \right)dt\]
\[ = 3\int t^{- \frac{1}{2}} dt - \int t^\frac{1}{2} dt\]
\[ = 3\left[ \frac{t^\frac{- 1}{2} + 1}{- \frac{1}{2} + 1} \right] - \left[ \frac{t^\frac{1}{2} + 1}{\frac{1}{2} + 1} \right] + C\]
\[ = 6\sqrt{t} - \frac{2}{3} t^\frac{3}{2} + C\]
\[ = 2\sqrt{t} \left( 3 - \frac{t}{3} \right) + C\]
\[ = 2\sqrt{t}\left( \frac{9 - t}{3} \right) + C \left[ \because t = 1 + 3x \right]\]
\[ = \frac{2}{3}\sqrt{1 + 3x} \left\{ \frac{9 - \left( 1 + 3x \right)}{3} \right\} + C\]
\[ = \frac{2}{3 \times 3}\sqrt{1 + 3x} \left( 8 - 3x \right) + C\]
\[ = \frac{2}{9}\left( 8 - 3x \right) \sqrt{1 + 3x} + C\]