integral of (x^2+x+1)/x

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

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∫ x²+x+1x dx

Solution

x²2 +x+ln|x|+C

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∫ x²+x+1x dx

Expand x²+x+1x : x+1+1x

=∫ x+1+1x dx

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

=∫ xdx+∫ 1dx+∫ 1x dx

∫ xdx=x²2

∫ 1dx=x

∫ 1x dx=ln|x|

=x²2 +x+ln|x|

Add a constant to the solution

=x²2 +x+ln|x|+C

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integral of (x^2+x+1)/x

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