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What is the integral of #int (2x-sinx) / (x^2 + cosx)#?

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1 Answer

Mahek ☮ · Kalit Gautam

Apr 17, 2018

Let #x^2 + cosx=t#

#(2x - sinx) dx = dt #

Replace in the original question.
#int (2x-sinx) / (x^2 + cosx)dx = int dt/t#

Again, replace.
#=>log|t| +c = log|x^2 + cosx| + c#

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