Question

Prove that $\int (x-2)\sqrt{2x^2-6x+5}dx=\frac{x-3}{2}\sqrt{{(x-\frac{3}{2})}^2+{\left(\frac{1}{2}\right)}^2}$ $+\frac{1}{2}{\left(\frac{1}{2}\right)}^2\log [(x-\frac{3}{2})+\sqrt{{(x-\frac{3}{2})}^2+{\left(\frac{1}{2}\right)}^2}].$

Answer 1

Let $I=\int (x-2)\sqrt{2x^2-6x+5}$
Substitute $x-2=l(4x-6)+m$
On comparing we get $4l=1,6l+m=-2$
Therefore
$l=\frac{1}{4},m=-\frac{1}{2}$
Hence
$I=l\int (4x-6)\sqrt{2x^2-6x+5}dx+m\int \sqrt{2x^2-6x+5}dx$
Substitute $2x^2-6x+5=t\Rightarrow (4x-6)dx=dt$
$I=l\int \sqrt{t}dt+m\sqrt{2}\int \sqrt{x^2-3x+5/2}dx$
$=l.\frac{2}{3}{t}^{3/2}+m\sqrt{2}\int \sqrt{{(x-\frac{3}{2})}^2+{\left(\frac{1}{2}\right)}^2}dx$
$=\frac{1}{4}.\frac{2}{3}{(2x^2-6x+5)}^{3/2}-\frac{1}{2}\sqrt{2I_1}=\frac{1}{6}{(2x^2-6x+5)}^{3/2}-\frac{1}{\sqrt{2}}{I}_1$
Where $I_1=\frac{x-\frac{3}{2}}{2}\sqrt{(x-\frac{3}{2})+{\left(\frac{1}{2}\right)}^2}$
Hence
$I=\frac{x-3}{2}\sqrt{{(x-\frac{3}{2})}^2+{\left(\frac{1}{2}\right)}^2}+\frac{1}{2}{\left(\frac{1}{2}\right)}^2\log [(x-\frac{3}{2})+\sqrt{{(x-\frac{3}{2})}^2+{\left(\frac{1}{2}\right)}^2}]$