How do you integrate $int (2x-sinx) / (x^2 + cosx)$ ?

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Answer 1 · 2016-01-31T19:24:30

The numerator is the derivative of the denominator.

Use substitution:

Let $u=x^2+cosx$ , which makes $du = (2x-sinx) dx$

So the integral is $int (du)/u = ln absu +C$ .

$int (2x-sinx)/(x^2+cosx) dx = ln abs(x^2+cosx) +C$

$= ln(x^2+cosx)+C$ $" "$ (since $x^2+cosx > 0$ for all $x$ )

How do you integrate int (2x-sinx) / (x^2 + cosx)int (2x-sinx) / (x^2 + cosx)?