Question

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

Answer 1

The numerator is the derivative of the denominator.

Explanation:

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`)