Question

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

Answer 1

$\mathrm{Let}I=\int \left(\frac{x-1}{\sqrt{x+4}}\right)dx$

Putting x + 4 = t
Then, x = t – 4
Difference both sides
dx = dt
Now integral becomes,
$$\begin{gathered} I=\int \left(\frac{t-4-1}{\sqrt{t}}\right)dt \\ =\int \left(\frac{t}{\sqrt{t}}-\frac{5}{\sqrt{t}}\right)dt \\ =\int \left(t^{\frac{1}{2}}-5t^{-\frac{1}{2}}\right)dt \\ =\frac{t^{{\displaystyle \frac{1}{2}+1}}}{{\displaystyle \frac{1}{2}+1}}-5\frac{t^{-{\displaystyle \frac{1}{2}}+1}}{-{\displaystyle \frac{1}{2}}+1}+\mathrm{C} \\ =\frac{2}{3}{t}^{\frac{3}{2}}-10\sqrt{t}+\mathrm{C} \\ =\frac{2}{3}{\left(x+4\right)}^{\frac{3}{2}}-10{\left(x+4\right)}^{\frac{1}{2}}+\mathrm{C} \end{gathered}$$