Mathematics

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


SOLUTION
$$I=\displaystyle \int (3x+2)\sqrt {2x^2-x+1}dx$$
for such questions write linear term as the first derivative of quadrant term.
i.e., $$3x-2=A\left (\dfrac {d}{dx} (2x^2-x+1) \right)+B$$
$$3x-2=A[4x-1]+B$$
$$3x-2=4Ax+B-A$$
$$4A=3$$ and $$B-A=-2$$
$$A=\dfrac {3}{4}\quad B-=\dfrac {3}{4}=-2$$
$$B==\dfrac {3}{4}-2==\dfrac {-5}{4}  $$
$$I=\displaystyle \int \left [\dfrac {3}{4}(4x-1)+\left (=\dfrac {-5}{4}\right)\right] \sqrt {2x^2-x+1}dx$$
$$=\dfrac {3}{4} \displaystyle \int \sqrt {2x^2-x+1}(4x-1)dx- \dfrac {5}{4} \displaystyle \int \sqrt {2x^2-x+1}dx$$
$$2x^2-x+1=t^2$$
$$(4x-1)dx=2td t$$
$$=\dfrac {3}{2}\displaystyle \int t^2 dt -=\dfrac {56}{4} \displaystyle \int \sqrt {\left (x-\dfrac {1}{4}\right)^2 +\left (\dfrac {\sqrt {15}}{4}\right)^2}$$
$$=\dfrac {3}{2}\left [\dfrac {t^3}{3}\right] \dfrac {-5\sqrt 2}{4} \left [\dfrac {\left (x-\dfrac {1}{4}\right)}{2} \sqrt {2x^2-x+1} \dfrac {-15}{32} \log \left |x-\dfrac {1}{4} + \sqrt {2x^2-x+1}\right | \right]$$
$$\Rightarrow \ \dfrac {(2x^2-x+1)^{3/2}}{2}\dfrac {-5\sqrt {2}}{4} \left [\left (\dfrac {x}{2}-\dfrac {1}{8}\right) \sqrt {2x^2-x+1} \dfrac {-15}{32} \log \left |x-\dfrac {1}{4} \right|+\sqrt {2x^2-x+1} \right]$$

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Subjective Medium Published on 17th 09, 2020
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