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- Integrate using basic integrals
- Integrate by partial fractions
- Integrate by substitution
- Integrate by parts
- Integrate using tabular integration
- Integrate by trigonometric substitution
- Weierstrass Substitution
- Integrate using trigonometric identities
- Product of Binomials with Common Term
- FOIL Method
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Since the integral $\int_{0}^{4}\frac{1}{x^2-2x-3}dx$ has a discontinuity inside the interval, we have to split it in two integrals
Learn how to solve definite integrals problems step by step online.
$\int_{0}^{3}\frac{1}{x^2-2x-3}dx+\int_{3}^{4}\frac{1}{x^2-2x-3}dx$
Learn how to solve definite integrals problems step by step online. Integrate the function 1/(x^2-2x+-3) from 0 to 4. Since the integral \int_{0}^{4}\frac{1}{x^2-2x-3}dx has a discontinuity inside the interval, we have to split it in two integrals. The integral \int_{0}^{3}\frac{1}{x^2-2x-3}dx results in: -0.3465736+\lim_{c\to0}\left(- \infty \right). Gather the results of all integrals. The integral \int_{3}^{4}\frac{1}{x^2-2x-3}dx results in: -0.0557859+\lim_{c\to3}\left(-\frac{1}{4}\ln\left(c-3\right)\right).