I used symbolab to work through the solution which turns out to be:\[\frac{1}{9}(\sqrt{3}\pi-\ln{(8)})\]The first step in the solution is to factorise the denominator. This is easy because \(x^3+1 \) is a sum of two cubes and so can be written as \((x+1)(x^2-x+1)\). Using partial fractions, the integral can be converted into the following form:\[ \frac{1}{3} \int\limits_0^1 \frac{x+1}{x^2-x+1} \text{d}x-\frac{1}{3} \int\limits_0^1 \frac{1}{x+1} \text{d}x \]The RHS integral is straightforward but the LHS one poses more difficulties. The way forward with the former is to write the quadratic in the denominator as a sum of two squares:\[ \frac{1}{3} \int\limits_0^1 \frac{x+1}{(x-\frac{1}{2})^2+\frac{3}{4}} \text{d}x \]This newly changed denominator can then be divided into the two parts of the numerator to yield:\[ \frac{1}{3} \int\limits_0^1 \frac{x}{(x-\frac{1}{2})^2+\frac{3}{4}}+\frac{1}{3} \int\limits_0^1 \frac{1}{(x-\frac{1}{2})^2+\frac{3}{4}} \]I'm not going to go into the details here but, with the appropriate substitutions, the LHS of this integral can be integrated to a logarithm and the RHS to an arctan using the formulae:\[ \int \frac{2t}{t^2+1} \text{d}t = \ln|t^2+1| + C \text{ and }\int \frac{1}{t^2+1} \text{d}t=\tan \!^{-1}{t}+C \]The indefinite integral for \( x/(x^3+1) \) turns out to be: \[ \frac{1}{6} \Big( \ln|4x^2-4x+4|+2\sqrt{3} \, \tan^{-1} \big( \frac{2}{\sqrt{3}}(x-\frac{1}{2}) \big) \Big)-\frac{1}{3}\ln|x+1|+C \]The definite integral for the limits of 0 and 1 is what was quoted earlier at the beginning of this post. In conclusion, I must say that Blogger is still inconsistently handling the display mode for LaTeX: sometimes it renders the LaTeX using the Javascript and other times it returns an image of the rendered code. Annoying.
No comments:
Post a Comment