Question

\[\int\frac{x - 1}{\sqrt{x + 4}} dx\]

Answer 1

\[\text{Let I} = \int\left( \frac{x - 1}{\sqrt{x + 4}} \right)dx\]

Putting  x + 4 = t
Then, x = t – 4
Difference both sides
dx = dt
Now integral becomes,

\[I = \int\left( \frac{t - 4 - 1}{\sqrt{t}} \right)dt\]
\[ = \int\left( \frac{t}{\sqrt{t}} - \frac{5}{\sqrt{t}} \right)dt\]
\[ = \int\left( t^\frac{1}{2} - 5 t^{- \frac{1}{2}} \right)dt\]
\[ = \frac{t^\frac{1}{2} + 1}{\frac{1}{2} + 1} - 5\frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} + C\]
\[ = \frac{2}{3} t^\frac{3}{2} - 10\sqrt{t} + C\]
\[ = \frac{2}{3} \left( x + 4 \right)^\frac{3}{2} - 10 \left( x + 4 \right)^\frac{1}{2} + C\]