Evaluate the integral: I = integral (x+1)/(x³(x²+1)²) dx
Evaluate the integral: $$ I = \int \frac{x+1}{x^3\left(x^2+1\right)^2}\,dx. $$
Explanation of Method
Decomposing into partial fractions over the repeated linear factor $x^3$ and the repeated irreducible quadratic factor $(x^2+1)^2$ reduces the integral to logarithms, power-rule terms, and arctangents via the standard reduction formula for $\int dx/(x^2+1)^2$.
- Set up the partial fraction decomposition.
- Solve for the coefficients by matching polynomial coefficients.
- Integrate each term.
Step 1 · Set up partial fractions
$$ \frac{x+1}{x^3(x^2+1)^2} = \frac Ax + \frac{B}{x^2} + \frac{C}{x^3} + \frac{Dx+E}{x^2+1} + \frac{Fx+G}{(x^2+1)^2}. $$
Multiplying through by $x^3(x^2+1)^2$:
$$ x+1 = Ax^2(x^2+1)^2 + Bx(x^2+1)^2 + C(x^2+1)^2 + (Dx+E)x^3(x^2+1) + (Fx+G)x^3. $$
Step 2 · Solve for the coefficients
Expanding the right side and matching coefficients by degree of $x$ (the left side is $0\cdot x^7+\cdots+0\cdot x^2+1\cdot x+1$):
$$ x^0:\ C=1 \qquad x^1:\ B=1 \qquad x^2:\ A+2C=0\Rightarrow A=-2 $$
$$ x^6:\ A+D=0\Rightarrow D=2 \qquad x^5:\ B+E=0\Rightarrow E=-1 $$
$$ x^3:\ 2B+E+G=0\Rightarrow G=-1 \qquad x^4:\ 2A+C+D+F=0\Rightarrow F=1 $$
Step 3 · Integrate the power-of-$x$ terms
$$ \int\left(\frac{-2}x+\frac1{x^2}+\frac1{x^3}\right)dx = -2\ln|x| - \frac1x - \frac1{2x^2}. $$
Step 4 · Integrate the $(x^2+1)$ term
$$ \int\frac{2x-1}{x^2+1}\,dx = \ln\left(x^2+1\right) - \arctan x. $$
Step 5 · Integrate the $(x^2+1)^2$ term
Using $\int\dfrac{x\,dx}{(x^2+1)^2}=-\dfrac1{2(x^2+1)}$ and the reduction formula $\int\dfrac{dx}{(x^2+1)^2}=\dfrac{x}{2(x^2+1)}+\dfrac12\arctan x$:
$$ \int\frac{x-1}{(x^2+1)^2}\,dx = -\frac1{2(x^2+1)} - \frac{x}{2(x^2+1)} - \frac12\arctan x = -\frac{x+1}{2(x^2+1)} - \frac12\arctan x. $$
Step 6 · Combine everything
$$ I = -2\ln|x| - \frac1x - \frac1{2x^2} + \ln\left(x^2+1\right) - \arctan x - \frac{x+1}{2(x^2+1)} - \frac12\arctan x + C. $$
Summary Table
| Coefficient | Value |
|---|---|
| $A$ | $-2$ |
| $B$ | $1$ |
| $C$ | $1$ |
| $D$ | $2$ |
| $E$ | $-1$ |
| $F$ | $1$ |
| $G$ | $-1$ |
Additional Notes
- Correction: the source correctly derived the coefficients $D=2,E=-1,F=1,G=-1$ in an earlier step, then second-guessed itself and substituted different, incorrect values into the final integration step, producing a wrong arctangent coefficient and a wrong sign on the $(x+1)/[2(x^2+1)]$ term. The derivation above re-solves the coefficient system from scratch (matching all seven powers of $x$, not just a subset) and confirms $D=2,E=-1,F=1,G=-1$ are indeed correct.
- The domain excludes $x=0$, where the denominator vanishes.
- The result can be checked by differentiating $\ln\left(\tfrac{x^2+1}{x^2}\right)-\tfrac1x-\tfrac1{2x^2}-\tfrac{x+1}{2(x^2+1)}-\tfrac32\arctan x$: combining all terms over the common denominator $x^3(x^2+1)^2$ and simplifying the numerator collapses every term of degree 2 through 6 to zero, leaving exactly $x+1$, matching the integrand.
Result
The solution to the integral is: