Final answer to the problem
Step-by-step Solution
Specify the solving method
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 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 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: -\frac{96}{277}+\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: -\frac{13}{233}+\lim_{c\to3}\left(-\frac{1}{4}\ln\left(c-3\right)\right).