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What is the antiderivative of #x/(x^2 + 4)#?

1 Answer

Jun 10, 2018

#int (xdx)/(x^2+4) = 1/2ln(x^2+4) + C#

Explanation:

Substitute #t=x^2+4#, #dt=2xdx#, so that:

#int (xdx)/(x^2+4) = 1/2 int dt/t = 1/2lnabst + C#

and undoing the substitution:

#int (xdx)/(x^2+4) = 1/2ln(x^2+4) + C#

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