Mathematics

# Evaluate: $\displaystyle \int\left (\dfrac{3x-1}{4x^{2}}-4x+17\right)dx$

##### SOLUTION
$\displaystyle\int\left(\dfrac{3x-1}{4x^2}-4x+17\right)dx$
$=\displaystyle\int\left(\dfrac{3}{4x}-\dfrac{1}{4x^2}-4x+17\right)dx$
$=\dfrac{3}{4}\displaystyle\int\dfrac{dx}{x}-\dfrac{1}{4}\displaystyle\int x^{-2}dx-4\displaystyle\int xdx+17\displaystyle\int dx$
$=\dfrac{3}{4}ln x-\dfrac{1}{4}\dfrac{x^{-2+1}}{-2+1}-\dfrac{4x^2}{2}+17x+c$
$=\dfrac{3}{4}ln x+\dfrac{1}{4x}-2x^2+17x+c$
Where c is the constant of integration.

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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