Question

Calculate each of the integrals in Exercises 17–46. For some integrals you may need to use polynomial long division, partial fractions, factoring or expanding, or the method of completing the square.

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

Short answer

The value is$2\ln |x-1|-\ln |x^2+x+1|+C$

Step-by-step solution

Step 1. Given Information  

The given integral is$\int \frac{3+3x}{x^3-1}dx$

Step 2. Calculation   

The partial fraction is calculated below,

$$\begin{gathered} \frac{x+1}{(x^3-1)}=\frac{x+1}{(x-1)(x^2+x+1)} \\ \frac{x+1}{(x^3-1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1} \\ \frac{x+1}{(x^3-1)}=\frac{\left(x-1\right)(Bx+C)+\left(x^2+x+1\right)A}{\left(x-1\right)(x^2+x+1)} \\ x+1=\left(x-1\right)(Bx+C)+\left(x^2+x+1\right)A \\ x+1=x^2(A+B)+x(A-B+C)+A-C \\ So, \\ A+B=0 \\ A-B+C=1 \\ A-C=1 \\ \mathrm{On}\mathrm{solving},\mathrm{we}\mathrm{get}, \\ A=\frac{2}{3},B=-\frac{2}{3},C=-\frac{1}{3} \end{gathered}$$

Step 3. Calculation 

The partial fraction can be written as follows,

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

Now, the integration is solved below,

$$\begin{gathered} 3\int \frac{x+1}{x^3-1}dx=3\int \left(\frac{-2x-1}{3x^2+3x+3}+\frac{2}{3x-3}\right)dx \\ =-\int \frac{2x+1}{x^2+x+1}dx+2\ln |x-1| \\ =-\int \frac{1}{u}du+2\ln |x-1|[u=x^2+x+1,du=(2x+1\left)dx\right] \\ =2\ln |x-1|-\ln \left|u\right|+C \\ =2\ln |x-1|-\ln |x^2+x+1|+C \end{gathered}$$