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Question Number 154175 by daus last updated on 15/Sep/21

Commented by daus last updated on 15/Sep/21
integrate it
	Answered by EDWIN88 last updated on 15/Sep/21

	$\int \frac{32 x}{(2 x - 1) (2 x - 3) (2 x - 5)} d x$ $\frac{32 x}{(2 x - 1) (2 x - 3) (2 x - 5)} = \frac{a}{2 x - 1} + \frac{b}{2 x - 3} + \frac{c}{2 x - 5}$ $a = [\frac{32 x}{(2 x - 3) (2 x - 5)}]_{x = \frac{1}{2}} = \frac{16}{(- 2) (- 4)} = 2$ $b = {[\frac{32 x}{(2 x - 1) (2 x - 5)}]}_{x = \frac{3}{2}} = \frac{48}{(2) (- 2)} = - 12$ $c = {[\frac{32 x}{(2 x - 1) (2 x - 3)}]}_{x = \frac{5}{2}} = \frac{80}{(4) (2)} = 10$ $I = \int \frac{2}{2 x - 1} d x - \int \frac{12}{2 x - 3} d x + \int \frac{10}{2 x - 5} d x$ $I = \ln ∣ 2 x - 1 ∣ - 6 \ln ∣ 2 x - 3 ∣ + 5 \ln ∣ 2 x - 5 ∣ + c$ $I = \ln ∣ \frac{C (2 x - 1) {(2 x - 5)}^{5}}{{(2 x - 3)}^{6}} ∣ .$
	Commented by daus last updated on 15/Sep/21
	appreaciated
	Answered by peter frank last updated on 15/Sep/21

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