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How do you integrate #x/((ln3)(x^2 + 4))#?

1 Answer

May 2, 2015

#intx/((ln3)(x^2+4))dx#

#ln3# is a constant so this turns down to,

#=> 1/ln3intx/(x^2+4)dx =1/ln3 int1/2*(2x)/(x^2+4)dx=1/(2ln3)int(2x)/(x^2+4)dx#

The numerator is the derivative of the denominator so,

#=>1/(2ln3)lnabs(x^2+4) + C#

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