Question: \int \frac{x^4 + 1}{x^3 - x^2 + x - 1} \, dx
Select one:
a. -\ln\left|\frac{x-1}{\sqrt{x^2+1}}\right| + x + \frac{x^2}{2} - \arctan x + c
b. \frac{x^2}{2} - \frac{x-1}{1-x} + \ln|x-1| + \frac{x^2-1}{x+1} - \arctan x - \frac{1}{2} \ln(x^2+1) + c
c. more than one of the solution is right
d. none of the solution is right
e. \frac{x^2}{2} + \ln|x-1| + x - \frac{1}{2} \ln(x^2+1) + c
f. \frac{x^2}{2} - \frac{1}{2} \ln(x^2+1) + \arctan x + x + \ln|x-1| + c
1. **State the problem:** We want to evaluate the integral $$\int \frac{x^4 + 1}{x^3 - x^2 + x - 1} \, dx.$$
2. **Factor the denominator:** Notice the denominator can be factored by grouping:
$$x^3 - x^2 + x - 1 = (x^3 - x^2) + (x - 1) = x^2(x - 1) + 1(x - 1) = (x - 1)(x^2 + 1).$$
3. **Rewrite the integral:**
$$\int \frac{x^4 + 1}{(x - 1)(x^2 + 1)} \, dx.$$
4. **Perform polynomial division:** Since the numerator degree (4) is higher than denominator degree (3), divide:
Divide $x^4 + 1$ by $x^3 - x^2 + x - 1$.
- Leading term division: $x^4 \div x^3 = x$.
- Multiply divisor by $x$: $x(x^3 - x^2 + x - 1) = x^4 - x^3 + x^2 - x$.
- Subtract:
$$ (x^4 + 1) - (x^4 - x^3 + x^2 - x) = 0 + x^3 - x^2 + x + 1.$$
5. **Repeat division with remainder:**
Divide $x^3 - x^2 + x + 1$ by $x^3 - x^2 + x - 1$:
- Leading term division: $x^3 \div x^3 = 1$.
- Multiply divisor by $1$: $x^3 - x^2 + x - 1$.
- Subtract:
$$ (x^3 - x^2 + x + 1) - (x^3 - x^2 + x - 1) = 0 + 0 + 0 + 2 = 2.$$
6. **Result of division:**
$$\frac{x^4 + 1}{x^3 - x^2 + x - 1} = x + 1 + \frac{2}{x^3 - x^2 + x - 1} = x + 1 + \frac{2}{(x - 1)(x^2 + 1)}.$$
7. **Rewrite integral:**
$$\int \frac{x^4 + 1}{x^3 - x^2 + x - 1} \, dx = \int (x + 1) \, dx + 2 \int \frac{1}{(x - 1)(x^2 + 1)} \, dx.$$
8. **Integrate polynomial part:**
$$\int (x + 1) \, dx = \frac{x^2}{2} + x + C.$$
9. **Partial fraction decomposition:**
Set
$$\frac{1}{(x - 1)(x^2 + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 1}.$$
Multiply both sides by denominator:
$$1 = A(x^2 + 1) + (Bx + C)(x - 1).$$
10. **Expand right side:**
$$1 = A x^2 + A + B x^2 - B x + C x - C.$$
Group terms:
$$1 = (A + B) x^2 + (-B + C) x + (A - C).$$
11. **Equate coefficients:**
- Coefficient of $x^2$: $A + B = 0$
- Coefficient of $x$: $-B + C = 0$
- Constant term: $A - C = 1$
12. **Solve system:**
From $A + B = 0$, $B = -A$.
From $-B + C = 0$, $C = B = -A$.
From $A - C = 1$, substitute $C = -A$:
$$A - (-A) = 1 \Rightarrow 2A = 1 \Rightarrow A = \frac{1}{2}.$$
Then
$$B = -\frac{1}{2}, \quad C = -\frac{1}{2}.$$
13. **Rewrite partial fractions:**
$$\frac{1}{(x - 1)(x^2 + 1)} = \frac{1/2}{x - 1} + \frac{-\frac{1}{2} x - \frac{1}{2}}{x^2 + 1} = \frac{1}{2(x - 1)} - \frac{1}{2} \frac{x + 1}{x^2 + 1}.$$
14. **Integrate partial fractions:**
$$2 \int \frac{1}{(x - 1)(x^2 + 1)} \, dx = 2 \int \left( \frac{1}{2(x - 1)} - \frac{1}{2} \frac{x + 1}{x^2 + 1} \right) dx = \int \frac{1}{x - 1} \, dx - \int \frac{x + 1}{x^2 + 1} \, dx.$$
15. **Integrate each term:**
- $$\int \frac{1}{x - 1} \, dx = \ln|x - 1| + C.$$
- Split second integral:
$$\int \frac{x + 1}{x^2 + 1} \, dx = \int \frac{x}{x^2 + 1} \, dx + \int \frac{1}{x^2 + 1} \, dx.$$
16. **Integrate first part:**
Use substitution $u = x^2 + 1$, $du = 2x dx$, so
$$\int \frac{x}{x^2 + 1} \, dx = \frac{1}{2} \int \frac{2x}{u} \, dx = \frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln|x^2 + 1| + C.$$
17. **Integrate second part:**
$$\int \frac{1}{x^2 + 1} \, dx = \arctan x + C.$$
18. **Combine integrals:**
$$\int \frac{x + 1}{x^2 + 1} \, dx = \frac{1}{2} \ln|x^2 + 1| + \arctan x + C.$$
19. **Substitute back:**
$$- \int \frac{x + 1}{x^2 + 1} \, dx = - \frac{1}{2} \ln|x^2 + 1| - \arctan x + C.$$
20. **Final integral expression:**
$$\int \frac{x^4 + 1}{x^3 - x^2 + x - 1} \, dx = \frac{x^2}{2} + x + \ln|x - 1| - \frac{1}{2} \ln|x^2 + 1| - \arctan x + C.$$
21. **Compare with options:**
This matches option (e):
$$\frac{x^2}{2} + \ln|x-1| + x - \frac{1}{2} \ln(x^2+1) + c.$$