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.


1 Answer 1



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


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

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

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

  • $\begingroup$ And here I am with the antiderivate in terms of logarithms, sign functions, and the fresnel integral. This is much better! $\endgroup$
    – Dando18
    Jul 22, 2017 at 16:01

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