Question

Evaluate
$\int (x-5)\sqrt{x^2+x}.dx$

Answer 1

$\int (x-5)\sqrt{x^2+x}dx$

Let $x-5=m\frac{d}{dx}(x^2+x)+N$

$x-5=m(2x+1)+N$

$x-5=2xm+m+N$

compare

$2m=1$ $m+N=-5$

$m=\frac{1}{2}$ $N=-5-\frac{1}{2}$

$N=-\frac{11}{2}$

So $\int (x-5)\sqrt{x^2+x}dx=\frac{1}{2}\int (2x+1)\sqrt{x^2+x}dx-\frac{1}{2}\int \sqrt{x^2+x}dx$

Now $\frac{1}{2}\int (2x+1)\sqrt{x^2+x}dx$ Here $\int \sqrt{x^2+x}dx$

Let $x^2+x=t$ $=\int \sqrt{x^2+x+\frac{1}{4}-\frac{1}{4}}dx$

$dx(2x+1)=dt$ $=\int \sqrt{{(x+\frac{1}{2})}^2-{\left(\frac{1}{2}\right)}^2}$

$\frac{1}{2}\int \sqrt{t}dt$ $=\frac{(x+\frac{1}{2})}{2}\sqrt{{(x+\frac{1}{2})}^2-{\left(\frac{1}{2}\right)}^2}+\frac{1}{4}{\cos }^{-1}\left(\frac{x+\frac{1}{2}}{\frac{1}{2}}\right)+C$

$\frac{1}{2}\int t^{\frac{1}{2}}dt$ $=\frac{2x+1}{4}\sqrt{x^2+x}+\frac{1}{4}{\cos }^{-1}(2x+1)+C$

$\frac{1}{2}\frac{t^{\frac{3}{2}}}{\frac{3}{2}}$

$\frac{(x^2+x{)}^{\frac{3}{2}}}{3}$

$\frac{(x^2+x{)}^{\frac{3}{2}}}{3}+\frac{2x+1}{4}\sqrt{x^2+x}+\frac{1}{4}{\cos }^{-1}(2x+1)+C$.