Question

`∫   x    \sqrt{x + 2}     dx ` 

Answer 1

`text{Let I} = ∫   x  \sqrt{x + 2} dx `

Putting  x + 2 = t
Then, x = t – 2
Difference both sides
dx = dt
Now, integral becomes

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