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

1 Answer
Bill K.
Aug 10, 2015

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

Explanation:

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#

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