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Question

Evaluate: $\int (\frac{3 x - 1}{4 x^{2}} - 4 x + 17) d x$

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Solution

$\int (\frac{3 x - 1}{4 x^{2}} - 4 x + 17) d x$
$= \int (\frac{3}{4 x} - \frac{1}{4 x^{2}} - 4 x + 17) d x$
$= \frac{3}{4} \int \frac{d x}{x} - \frac{1}{4} \int x^{- 2} d x - 4 \int x d x + 17 \int d x$
$= \frac{3}{4} l n x - \frac{1}{4} \frac{x^{- 2 + 1}}{- 2 + 1} - \frac{4 x^{2}}{2} + 17 x + c$
$= \frac{3}{4} l n x + \frac{1}{4 x} - 2 x^{2} + 17 x + c$
Where c is the constant of integration.

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