How do you integrate $(8x)/(4x^2+1)$ ?

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Answer 1 · 2015-08-10T03:36:59

$int (8x)/(4x^2+1)\ dx=ln(4x^2+1)+C$

Do a substitution: let $u=4x^2+1$ so $du = 8x\ dx$ . Therefore,

$int (8x)/(4x^2+1)\ dx=int 1/u\ du=ln|u|+C$

$=ln|4x^2+1|+C$ .

Since $4x^2+1 geq 1 > 0$ for all $x in RR$ , we can get rid of the absolute value signs and write:

$int (8x)/(4x^2+1)\ dx=ln(4x^2+1)+C$

How do you integrate (8x)/(4x^2+1)(8x)/(4x^2+1)?