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invers 20((x-3)(x+4))/((x-3)^2(x-5))

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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∫ xx²−6x+8 dx

Solution

12 ln|x²−6x+8|−3(12 ln|x−2|−12 ln|x−4|)+C

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∫ xx²−6x+8 dx

Vervollständige das Quadrat x²−6x+8: (x−3)²−1

=∫ x(x−3)²−1 dx

Wende U-Substitution an: ∫ u+3u²−1 du

=∫ u+3u²−1 du

Multipliziere aus u+3u²−1 : uu²−1 +3u²−1

Wende die Summenregel an: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx

=∫ uu²−1 du+∫ 3u²−1 du

∫ uu²−1 du=12 ln|u²−1|

∫ 3u²−1 du=−3(12 ln|u+1|−12 ln|u−1|)

=12 ln|u²−1|−3(12 ln|u+1|−12 ln|u−1|)

Setze in u=x−3 ein

=12 ln|(x−3)²−1|−3(12 ln|x−3+1|−12 ln|x−3−1|)

Vereinfache 12 ln|(x−3)²−1|−3(12 ln|x−3+1|−12 ln|x−3−1|): 12 ln|x²−6x+8|−3(12 ln|x−2|−12 ln|x−4|)

=12 ln|x²−6x+8|−3(12 ln|x−2|−12 ln|x−4|)

Füge eine Konstante zur Lösung hinzu

=12 ln|x²−6x+8|−3(12 ln|x−2|−12 ln|x−4|)+C

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

invers 20((x-3)(x+4))/((x-3)^2(x-5))

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