integral of (x+2)/(x^3+8)

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

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∫ x+2x³+8 dx

Solution

112 ln|x²−2x+4|−112 ln|13 (x−1)²+1|+1√3 arctan(1√3 (x−1))+C

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∫ x+2x³+8 dx

Expand x+2x³+8 : xx³+8 +2x³+8

Apply the Sum Rule: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx

=∫ xx³+8 dx+∫ 2x³+8 dx

∫ xx³+8 dx=−16 ln|x+2|+112 (ln|x²−2x+4|+2√3arctan(1√3 (x−1)))

∫ 2x³+8 dx=2(112 ln|x+2|+112 (−12 ln|13 (x−1)²+1|+√3arctan(1√3 (x−1))))

=−16 ln|x+2|+112 (ln|x²−2x+4|+2√3arctan(1√3 (x−1)))+2(112 ln|x+2|+112 (−12 ln|13 (x−1)²+1|+√3arctan(1√3 (x−1))))

Simplify −16 ln|x+2|+112 (ln|x²−2x+4|+2√3arctan(1√3 (x−1)))+2(112 ln|x+2|+112 (−12 ln|13 (x−1)²+1|+√3arctan(1√3 (x−1)))): 112 ln|x²−2x+4|−112 ln|13 (x−1)²+1|+1√3 arctan(1√3 (x−1))

=112 ln|x²−2x+4|−112 ln|13 (x−1)²+1|+1√3 arctan(1√3 (x−1))

Add a constant to the solution

=112 ln|x²−2x+4|−112 ln|13 (x−1)²+1|+1√3 arctan(1√3 (x−1))+C

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

integral of (x+2)/(x^3+8)

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