# Definite integral problem /doubt

What is the integral $$\int_{-1}^{1} \frac{\sin x - x^2}{3-|x|}\ dx$$ I have tried splitting the integral at $0$ and then separating the denominator.

Let

$$I=\int_{-1}^{1} \frac{\sin x - x^2}{3-|x|}\ dx$$

Then

$$I=\int_{-1}^{1} \frac{\sin x}{3-|x|}\ dx-\int_{-1}^{1} \frac{x^2}{3-|x|}\ dx$$

Notice the first term is odd while the second is even. Hence

$$I=-2\int_{0}^{1} \frac{x^2}{3-x}\ dx$$

• And here I am with the antiderivate in terms of logarithms, sign functions, and the fresnel integral. This is much better! Jul 22, 2017 at 16:01