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$\int_{-5}^{5}\frac{4x^4-2x^2+4}{x}dx$
Learn how to solve problems step by step online. Integrate the function (8x^4-4x^2+8)/(2x) from -5 to 5. Simplify the expression inside the integral. Since the integral \int_{-5}^{5}\frac{4x^4-2x^2+4}{x}dx has a discontinuity inside the interval, we have to split it in two integrals. The integral \int_{-5}^{0}\frac{4x^4-2x^2+4}{x}dx results in: \int_{-5}^{0}\frac{4x^4-2x^2+4}{x}dx+\int_{0}^{0}\frac{4x^4-2x^2+4}{x}dx. The integral \int_{-5}^{0}\frac{4x^4-2x^2+4}{x}dx results in: \int_{-5}^{0}\frac{4x^4-2x^2+4}{x}dx+\int_{0}^{0}\frac{4x^4-2x^2+4}{x}dx.