Evaluate the integral: I = integral (x³+1)/(x(x-1)³) dx

Evaluate the integral: $$ I = \int \frac{x^3+1}{x(x-1)^3}\,dx. $$

Step 1 · Set up partial fractions

$$ \frac{x^3+1}{x(x-1)^3} = \frac Ax + \frac{B}{x-1} + \frac{C}{(x-1)^2} + \frac{D}{(x-1)^3}. $$

Multiplying through: $x^3+1=A(x-1)^3+Bx(x-1)^2+Cx(x-1)+Dx$.

Step 2 · Solve for the coefficients

Setting $x=0$: $1=-A$, so $A=-1$. Setting $x=1$: $2=D$. Matching the $x^3$ and $x^2$ coefficients from the expansion gives $B=2$ and $C=1$.

Step 3 · Integrate each term

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

Using $\int(x-1)^{-2}dx=-(x-1)^{-1}$ and $\int(x-1)^{-3}dx=-\tfrac12(x-1)^{-2}$:

$$ I = -\ln|x| + 2\ln|x-1| - \frac{1}{x-1} - \frac{1}{(x-1)^2} + C. $$

Summary Table

Additional Notes

• The domain excludes $x=0$ and $x=1$, the roots of the denominator.
• This problem has the same repeated-factor structure as Problem 243, but with a cubic numerator; the $D=2$ coefficient on $(x-1)^{-3}$ correctly integrates here to $-1/(x-1)^2$ (note the necessary minus sign from the power rule on a negative exponent, which is easy to drop, as happened in Problem 243).
• The result can be checked by differentiating $-\ln|x|+2\ln|x-1|-\tfrac{1}{x-1}-\tfrac{1}{(x-1)^2}$: this gives $-\tfrac1x+\tfrac{2}{x-1}+\tfrac{1}{(x-1)^2}+\tfrac{2}{(x-1)^3}$, matching the partial-fraction decomposition (and hence the integrand) term by term.