Chapter G, Problem 29E

To determine

To calculate:

Integral of following function.

  ${\displaystyle \int \frac{1}{x^3-1}dx}$ .

Answer to Problem 29E

Integral of given function is $1/3\ln \left|x-1\right|-1/6\ln \left|{\left(x+1/2\right)}^2+3/4\right|-1/2\left(2/\sqrt{3}\right){\tan }^{-1}\frac{2x+1}{\sqrt{3}}+C$ .

Explanation of Solution

Given information:

  ${\displaystyle \int \frac{1}{x^3-1}dx}$

Calculation:

Here given function is,

  ${\displaystyle \int \frac{1}{x^3-1}dx}$ .

The integrand ${\displaystyle \int \frac{1}{x^3-1}dx}$ we have to set up in partial fraction form,

  ${\displaystyle \int \frac{1}{x^3-1}={\displaystyle \int \frac{1}{\left(x-1\right)\left(x^2+x+1\right)}dx}}$

Now apply partial fraction decomposition to get

  $\begin{array}{l}\frac{1}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}\mathrm{...}\left(1\right)\\ 1=A\left(x^2+x+1\right)+\left(Bx+C\right)\left(x-1\right)\\ 1=A{x}^2+Ax+A+B{x}^2-Bx+Cx-C\\ 1=\left(A+B\right)x^2+\left(A-B+C\right)x+\left(A-C\right)\end{array}$

Comparing the coefficients on both sides of above equation we have,

  $\begin{array}{l}0=A+B\\ 0=A-B+C\\ 1=A-C\end{array}$

Thus we get following coefficients by solving,

  $\begin{array}{l}A=\frac{1}{3}\\ B=\frac{-1}{3}\\ C=\frac{-2}{3}\end{array}$

Plug the values of $A,B,C$ in eq. (1),

  $\begin{array}{l}\frac{1}{\left(x-1\right)\left(x^2+x+1\right)}={\displaystyle \int \frac{1/3}{x-1}}-\frac{1/3x}{x^2+x+1}-\frac{2/3}{x^2+x+1}dx\\ ={\displaystyle \int \frac{1/3}{x-1}}-\frac{1/3x}{{\left(x+1/2\right)}^2+3/4}-\frac{2/3}{{\left(x+1/2\right)}^2+3/4}dx\\ ={\displaystyle \int \frac{1/3}{x-1}}-\frac{1/3\left(x+1/2\right)}{{\left(x+1/2\right)}^2+3/4}-\frac{1/2}{{\left(x+1/2\right)}^2+3/4}dx\\ =1/3\ln \left|x-1\right|-1/6\ln \left|{\left(x+1/2\right)}^2+3/4\right|-1/2\left(2/\sqrt{3}\right){\tan }^{-1}\frac{2x+1}{\sqrt{3}}+C\end{array}$

Thus the given integral becomes,

  $=1/3\ln \left|x-1\right|-1/6\ln \left|{\left(x+1/2\right)}^2+3/4\right|-1/2\left(2/\sqrt{3}\right){\tan }^{-1}\frac{2x+1}{\sqrt{3}}+C$