Question

Solve $\int \left(3x-2\right)\sqrt{2x^2-x+1}dx$

Answer 1

$I=\int (3x+2)\sqrt{2x^2-x+1}dx$

for such questions write linear term as the first derivative of quadrant term.

i.e., $3x-2=A\left(\frac{d}{dx}\right(2x^2-x+1\left)\right)+B$

$3x-2=A[4x-1]+B$

$3x-2=4Ax+B-A$

$4A=3$ and $B-A=-2$

$A=\frac{3}{4}B-=\frac{3}{4}=-2$

$B==\frac{3}{4}-2==\frac{-5}{4}$

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

$=\frac{3}{4}\int \sqrt{2x^2-x+1}(4x-1)dx-\frac{5}{4}\int \sqrt{2x^2-x+1}dx$

$2x^2-x+1=t^2$

$(4x-1)dx=2tdt$

$=\frac{3}{2}\int t^{2}dt-=\frac{56}{4}\int \sqrt{{(x-\frac{1}{4})}^2+{\left(\frac{\sqrt{15}}{4}\right)}^2}$

$=\frac{3}{2}\left[\frac{t^3}{3}\right]\frac{-5\sqrt{2}}{4}\left[\frac{(x-\frac{1}{4})}{2}\sqrt{2x^2-x+1}\frac{-15}{32}\log |x-\frac{1}{4}+\sqrt{2x^2-x+1}|\right]$

$\Rightarrow \frac{(2x^2-x+1{)}^{3/2}}{2}\frac{-5\sqrt{2}}{4}[(\frac{x}{2}-\frac{1}{8})\sqrt{2x^2-x+1}\frac{-15}{32}\log |x-\frac{1}{4}|+\sqrt{2x^2-x+1}]$