Question

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

Answer 1

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

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

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

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

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

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

$=\frac{3}{4}(2x^2-x+1)-\frac{5}{2\sqrt{2}}\int \sqrt{x^2-2\times x\times \frac{1}{4}+\frac{1}{16}+\frac{1}{2}-\frac{1}{16}}dx$

$=\frac{3}{4}(2x^2-x+1)-\frac{5}{2\sqrt{2}}\int \sqrt{\left(x\frac{-1}{4}\right){\left(\frac{\sqrt{7}}{4}\right)}^2}dx$

$=\frac{3}{4}(2x^2-x+1)-\frac{5}{2\sqrt{2}}[\frac{1}{2}\times (x-\frac{1}{4})\sqrt{x^2-\frac{x}{2}+\frac{1}{2}}+\frac{1}{2}\times \frac{7}{16}\times {\sin }^{-1}\left(\frac{4x-1}{\sqrt{7}}\right)]+C$