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- Integrate by substitution
- Integrate by partial fractions
- Integrate by parts
- Integrate using tabular integration
- Integrate by trigonometric substitution
- Weierstrass Substitution
- Integrate using trigonometric identities
- Integrate using basic integrals
- Product of Binomials with Common Term
- FOIL Method
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Since the integral $\int_{-2}^{1}\frac{x+2}{2x^2-4x}dx$ has a discontinuity inside the interval, we have to split it in two integrals
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$\int_{-2}^{0}\frac{x+2}{2x^2-4x}dx+\int_{0}^{1}\frac{x+2}{2x^2-4x}dx$
Learn how to solve problems step by step online. Integrate the function (x+2)/(2x^2-4x) from -2 to 1. Since the integral \int_{-2}^{1}\frac{x+2}{2x^2-4x}dx has a discontinuity inside the interval, we have to split it in two integrals. The integral \int_{-2}^{0}\frac{x+2}{2x^2-4x}dx results in: \int_{-2}^{0}\frac{x}{2x^2-4x}dx+\int_{-2}^{0}\frac{1}{x^2-2x}dx. Gather the results of all integrals. The integral \int_{0}^{1}\frac{x+2}{2x^2-4x}dx results in: \int_{0}^{1}\frac{x}{2x^2-4x}dx+\int_{0}^{1}\frac{1}{x^2-2x}dx.
