integral of x/(x^2+1)

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

☐²

x^☐

√☐

^□√☐

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log_□

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ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

simplify solve for inverse tangent line

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∫ xx²+1 dx

Solution

12 ln|x²+1|+C

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∫ xx²+1 dx

Apply u-substitution: ∫ 12u du

=∫ 12u du

Take the constant out: ∫a·f(x)dx=a·∫f(x)dx

=12 · ∫ 1u du

Use the common integral: ∫ 1u du=ln(|u|)

=12 ln|u|

Substitute back u=x²+1

=12 ln|x²+1|

Add a constant to the solution

=12 ln|x²+1|+C

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

integral of x/(x^2+1)

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