derivative of (x^4^5)

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

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ddx

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

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∫ x+x³+x⁵1−x²+x⁴ dx

Solution

12√3 (√3x²+√3ln|−4x²+4x⁴+4|+2arctan(2x²−1√3 ))+C

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Solution steps

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∫ x+x³+x⁵1−x²+x⁴ dx

Apply u-substitution: ∫ 1+u+u²2(u²+1−u) du

=∫ 1+u+u²2(u²+1−u) du

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

=12 · ∫ 1+u+u²u²+1−u du

Expand 1+u+u²u²+1−u : 1u²+1−u +uu²+1−u +u²u²+1−u

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

=12 (∫ 1u²+1−u du+∫ uu²+1−u du+∫ u²u²+1−u du)

∫ 1u²+1−u du=2√3 arctan(2u−1√3 )

∫ uu²+1−u du=12 ln|4u²−4u+4|+1√3 arctan(2u−1√3 )

∫ u²u²+1−u du=12 ln|4u²−4u+4|−1√3 arctan(2u−1√3 )+u

=12 (2√3 arctan(2u−1√3 )+12 ln|4u²−4u+4|+1√3 arctan(2u−1√3 )+12 ln|4u²−4u+4|−1√3 arctan(2u−1√3 )+u)

Substitute back u=x²

=12 (2√3 arctan(2x²−1√3 )+12 ln|4(x²)²−4x²+4|+1√3 arctan(2x²−1√3 )+12 ln|4(x²)²−4x²+4|−1√3 arctan(2x²−1√3 )+x²)

Simplify 12 (2√3 arctan(2x²−1√3 )+12 ln|4(x²)²−4x²+4|+1√3 arctan(2x²−1√3 )+12 ln|4(x²)²−4x²+4|−1√3 arctan(2x²−1√3 )+x²): 12√3 (√3x²+√3ln|−4x²+4x⁴+4|+2arctan(2x²−1√3 ))

=12√3 (√3x²+√3ln|−4x²+4x⁴+4|+2arctan(2x²−1√3 ))

Add a constant to the solution

=12√3 (√3x²+√3ln|−4x²+4x⁴+4|+2arctan(2x²−1√3 ))+C

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

derivative of (x^4^5)

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