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

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

Step 1 · Set up partial fractions

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

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

Step 2 · Solve for the coefficients

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

Step 3 · Integrate each term

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

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

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

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

Summary Table

Additional Notes

• Correction: the source computed $\int dx/(x-1)^3=+\tfrac1{2(x-1)^2}$, dropping the minus sign that the power rule requires for a negative exponent ($\int(x-1)^{-3}dx=(x-1)^{-2}/(-2)$). Since the term in the decomposition has coefficient $D=2$, this sign flip carried through to give $+\tfrac{1}{(x-1)^2}$ in the source's final answer instead of the correct $-\tfrac{1}{(x-1)^2}$.
• The domain excludes $x=0$ and $x=1$, the roots of the denominator.
• The result can be checked by differentiating $-\ln|x|+\ln|x-1|+\tfrac{1}{x-1}-\tfrac{1}{(x-1)^2}$: this gives $-\tfrac1x+\tfrac{1}{x-1}-\tfrac{1}{(x-1)^2}+\tfrac{2}{(x-1)^3}$, matching the partial-fraction decomposition (and hence the integrand) term by term.