Question

If $\int (x+2)\sqrt{x+1}dx$ is equal to $\frac{2}{p}(x+1{)}^{\frac{3}{2}}(qx+8)+C,$ then the value of $2(p-q)$ is equal to

Answer 1

Let $I=\int (x+2)\sqrt{x+1}dx$
Put $x+1=t^2\Rightarrow dx=2tdt$
$\Rightarrow I=\int (t^2+1)t\cdot 2tdt$
$=2\int (t^4+t^2)dt$
$=2(\frac{t^5}{5}+\frac{t^3}{3})+C$
$=\frac{2}{15}{t}^3(3t^2+5)+C$
$=\frac{2}{15}(x+1{)}^{\frac{3}{2}}(3x+8)+C$
$\therefore p=15,q=3$
Hence, $2(p-q)=2(15-3)=24$