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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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