derivative of tan(x-x)

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simplify solve for inverse tangent line

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∫ √4+x²dx

Solution

4(18 x√4+x²+12 ln(|x+√4+x²|2 ))+C

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∫ √4+x²dx

Apply Trigonometric Substitution: ∫ 4sec³(u)du

=∫ 4sec³(u)du

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

=4· ∫ sec³(u)du

Apply Integral Reduction

=4(sec²(u)sin(u)2 +12 · ∫ sec(u)du)

∫ sec(u)du=ln|tan(u)+sec(u)|

=4(sec²(u)sin(u)2 +12 ln|tan(u)+sec(u)|)

Substitute back u=arctan(12 x)

=4(sec²(arctan(12 x))sin(arctan(12 x))2 +12 ln|tan(arctan(12 x))+sec(arctan(12 x))|)

Simplify 4(sec²(arctan(12 x))sin(arctan(12 x))2 +12 ln|tan(arctan(12 x))+sec(arctan(12 x))|): 4(18 x√4+x²+12 ln(|x+√4+x²|2 ))

=4(18 x√4+x²+12 ln(|x+√4+x²|2 ))

Add a constant to the solution

=4(18 x√4+x²+12 ln(|x+√4+x²|2 ))+C

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

derivative of tan(x-x)

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