∫_2 ^4 ((3x−2)/(x^2 −4)) dx = ?


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Question Number 94079 by Rio Michael last updated on 16/May/20

$\underset{2}{\int^{4}} \frac{3 x - 2}{x^{2} - 4} d x = ?$
	Answered by Kunal12588 last updated on 16/May/20

	$\frac{3}{2} [\int_{2}^{4} \frac{2 x}{x^{2} - 4} d x - \frac{4}{3} \int_{2}^{4} \frac{d x}{x^{2} - 4} d x]$ $= \frac{3}{2} {[l n ∣ x^{2} - 4 ∣ - \frac{1}{3} l n ∣ \frac{x - 2}{x + 2} ∣]}_{2}^{4}$ $= \frac{3}{2} {[l n ∣ \frac{(x^{2} - 4) {(x + 2)}^{1 / 3}}{{(x - 2)}^{1 / 3}} ∣]}_{2}^{4}$ $= \frac{3}{2} {[l n ∣ {(x - 2)}^{2 / 3} {(x + 2)}^{4 / 3} ∣]}_{2}^{4}$ $= {[l n ∣ (x - 2) {(x + 2)}^{2} ∣]}_{2}^{4}$ $o h h n o! {i t}^{'} s d i v e r g e n t$
	Answered by niroj last updated on 16/May/20
	$∫_2 ^( 4) (( 3x−2)/(x^2 −4))dx = [ 2∫ (1/(x+2))dx +∫(1/(x−2))dx]_2 ^4 = [2∫ (1/(x+2))dx +∫(1/(x−2))dx]_2 ^4 = [2log (x+2)+log(x−2)]_2 ^4 = [log(x+2)^2 +log(x−2)]_2 ^4 = [log {(x+2)^2 (x−2}]_2 ^4 = {log (4+2)^2 (4−2)−log (2+2)^2 (2−2)] = log 36.2−log16×0 = log 72 //. = 1.857$
	$\int_{2}^{4} \frac{3 x - 2}{x^{2} - 4} dx$ $= {[2 \int \frac{1}{x + 2} dx + \int \frac{1}{x - 2} dx]}_{2}^{4}$ $= {[2 \int \frac{1}{x + 2} dx + \int \frac{1}{x - 2} dx]}_{2}^{4}$ $= {[2 \log (x + 2) + \log (x - 2)]}_{2}^{4}$ $= {[\log {(x + 2)}^{2} + \log (x - 2)]}_{2}^{4}$ $= [\log {{{(x + 2)}^{2} (x - 2}]}_{2}^{4}$ $= {\log {(4 + 2)}^{2} (4 - 2) - \log {(2 + 2)}^{2} (2 - 2)]$ $= \log 36.2 - \log 16 \times 0$ $= \log 72 / / .$ $= 1.857$
	Commented by i jagooll last updated on 17/May/20

	$it should be \log (16 \times 0)$ $not \log 16 \times 0 sir$
	Commented by mathmax by abdo last updated on 17/May/20

	$e r r o r o f c a l c u l u s t h i s i n t e g r a l i s d i v e r g e n t . .!$
	Commented by Kunal12588 last updated on 16/May/20

	$c h e c k l a s t 5^{t h} l i n e$ $l o g ({(x + 2)}^{2} (x - 2))$
	Answered by Rio Michael last updated on 16/May/20

	$what i got is,$ $\frac{3 x - 2}{(x - 2) (x + 2)} = \frac{1}{x - 2} - \frac{2}{x - 2}$ $\Rightarrow \underset{2}{\int^{4}} \frac{3 x - 2}{x^{2} + 2} d x = \lim_{t \to 2} [\underset{t}{\int^{4}} (\frac{1}{x - 2} - \frac{2}{x + 2}) d x]$ $= \lim_{t \to 2} {[\ln (x - 2) - 2 \ln (x + 2)]}_{t}^{4}$ $= \lim_{t \to 2} [\ln 2 - 2 \ln 6 - (\ln (t - 2) - 2 \ln (t + 2)] = + \infty$ $diverges .$
	Commented by i jagooll last updated on 17/May/20

	$\frac{1}{x - 2} - \frac{2}{x + 2} = \frac{x + 2 - 2 (x - 2)}{(x + 2) (x - 2)}$ $\frac{x + 2 - 2 x + 4}{x^{2} - 4} = \frac{- x + 6}{x^{2} - 4} \neq \frac{3 x - 2}{x^{2} - 4}$
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