Corrected exercises: partial fractions (MATH 141)

a) The degrees are $2$ and $1$, so the fraction is improper and the division is compulsory. $x^2 + 3x$ divided by $x + 1$: first term $x$, and $x(x + 1) = x^2 + x$ leaves $2x$; next term $2$, and $2(x + 1) = 2x + 2$ leaves $-2$. So $x^2 + 3x = (x + 1)(x + 2) - 2$ and $\frac{x^2 + 3x}{x + 1} = x + 2 - \frac{2}{x + 1}$. The quotient $x + 2$ is the dashed line: as $x \to \pm\infty$ the proper part $-\frac{2}{x + 1}$ tends to $0$, so the curve hugs $y = x + 2$, its slant asymptote. For $x > -1$ the gap $-\frac{2}{x + 1}$ is negative, so the curve lies BELOW the line, as the figure shows. Integrating term by term: $\int \frac{x^2 + 3x}{x + 1}\,dx = \frac{x^2}{2} + 2x - 2\ln|x + 1| + C$. Check by differentiating: $x + 2 - \frac{2}{x + 1}$, which is the integrand.

b) Degrees $3$ and $2$: divide. $x^3 = x(x^2 - 4) + 4x$, so $\frac{x^3}{x^2 - 4} = x + \frac{4x}{x^2 - 4}$. The remainder is now proper, and before reaching for $A$ and $B$ it is worth reading its SHAPE: the numerator $4x$ is twice the derivative of $x^2 - 4$. The substitution $u = x^2 - 4$, $du = 2x\,dx$ gives $\int \frac{4x}{x^2 - 4}\,dx = 2\ln|x^2 - 4| + C$. So $\int \frac{x^3}{x^2 - 4}\,dx = \frac{x^2}{2} + 2\ln|x^2 - 4| + C$. Partial fractions give the same thing, $\frac{4x}{x^2 - 4} = \frac{2}{x - 2} + \frac{2}{x + 2}$, whose integral $2\ln|x - 2| + 2\ln|x + 2|$ is $2\ln|x^2 - 4|$: the decomposition is correct but unnecessary. The division was not optional; the decomposition was.

c) At $x = 0$ the left side is $\frac{1}{-2} = -\frac{1}{2}$, while the right side is $\frac{2/3}{-1} + \frac{-5/3}{2} = -\frac{2}{3} - \frac{5}{6} = -\frac{3}{2}$. The identity is false. The numerator and the denominator both have degree $2$, so the fraction tends to $1$ at infinity, while any sum $\frac{A}{x - 1} + \frac{B}{x + 2}$ tends to $0$: no choice of $A$ and $B$ can work. The cover-up did not warn him, because it reads the behaviour near each pole and is blind to the quotient: the constants it gives are right, but a whole term is missing. Correct route: $x^2 + 1 = 1 \cdot (x^2 + x - 2) + (-x + 3)$, so $\frac{x^2 + 1}{x^2 + x - 2} = 1 + \frac{-x + 3}{(x - 1)(x + 2)}$. Then $-x + 3 = A(x + 2) + B(x - 1)$; $x = 1$ gives $2 = 3A$, $A = \frac{2}{3}$; $x = -2$ gives $5 = -3B$, $B = -\frac{5}{3}$. Hence $\int \frac{x^2 + 1}{x^2 + x - 2}\,dx = x + \frac{2}{3}\ln|x - 1| - \frac{5}{3}\ln|x + 2| + C$. His error costs the term $x$, and on a definite integral it costs the whole length of the interval.

d) Degrees $2$ and $1$: divide. $2x^2 + x = (x + 1)(2x - 1) + 1$, since $(x + 1)(2x - 1) = 2x^2 + x - 1$. So $\frac{2x^2 + x}{x + 1} = 2x - 1 + \frac{1}{x + 1}$, and the pole $x = -1$ is outside $[0, 1]$, so the integrand is continuous there. $\int_0^1 \left(2x - 1 + \frac{1}{x + 1}\right)dx = \left[x^2 - x + \ln|x + 1|\right]_0^1 = (1 - 1 + \ln 2) - (0 - 0 + \ln 1) = \ln 2$. Sanity check: the integrand rises from $0$ at $x = 0$ to $\frac{3}{2}$ at $x = 1$, so an area of about $0.69$ is believable.

a) Degree $2$ over degree $3$: proper, no division. Factor: $x^3 - x = x(x - 1)(x + 1)$, three distinct linear factors, so the form is $\frac{6x^2 - x - 1}{x(x - 1)(x + 1)} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x + 1}$. Multiply both sides by $x(x - 1)(x + 1)$: $6x^2 - x - 1 = A(x - 1)(x + 1) + Bx(x + 1) + Cx(x - 1)$, an identity of polynomials, true for EVERY $x$. At $x = 0$ the $B$ and $C$ terms vanish: $-1 = -A$, so $A = 1$. At $x = 1$: $6 - 1 - 1 = 4 = 2B$, so $B = 2$. At $x = -1$: $6 + 1 - 1 = 6 = 2C$, so $C = 3$. Quick test: the coefficient of $x^2$ on the right is $A + B + C = 6$, which matches the left. Therefore $\int \frac{6x^2 - x - 1}{x^3 - x}\,dx = \ln|x| + 2\ln|x - 1| + 3\ln|x + 1| + K$, where $K$ is the constant of integration (the letter $C$ is taken). The step that earns the method mark is the polynomial identity: the values $x = 0, 1, -1$ are forbidden in the FRACTION but allowed in the identity, which is why the cover-up is legitimate.

b) $2x^2 + 5x - 3$ has discriminant $25 + 24 = 49 = 7^2$, so it factors: $(2x - 1)(x + 3)$. Form: $\frac{1}{(2x - 1)(x + 3)} = \frac{A}{2x - 1} + \frac{B}{x + 3}$, so $1 = A(x + 3) + B(2x - 1)$. At $x = \frac{1}{2}$: $1 = \frac{7}{2}A$, $A = \frac{2}{7}$. At $x = -3$: $1 = -7B$, $B = -\frac{1}{7}$. Now the trap of the chapter: $\int \frac{2/7}{2x - 1}\,dx$ is NOT $\frac{2}{7}\ln|2x - 1|$. With $u = 2x - 1$, $du = 2\,dx$, it is $\frac{2}{7} \cdot \frac{1}{2}\ln|2x - 1| = \frac{1}{7}\ln|2x - 1|$. So $\int \frac{dx}{2x^2 + 5x - 3} = \frac{1}{7}\ln|2x - 1| - \frac{1}{7}\ln|x + 3| + C = \frac{1}{7}\ln\left|\frac{2x - 1}{x + 3}\right| + C$. Differentiating the answer confirms it: $\frac{1}{7}\left(\frac{2}{2x - 1} - \frac{1}{x + 3}\right) = \frac{1}{7} \cdot \frac{2x + 6 - 2x + 1}{(2x - 1)(x + 3)} = \frac{1}{(2x - 1)(x + 3)}$.

c) $\frac{1}{x^2 - 1} = \frac{1}{(x - 1)(x + 1)} = \frac{1/2}{x - 1} - \frac{1/2}{x + 1}$ (cover-up at $x = 1$ gives $\frac{1}{2}$, at $x = -1$ gives $-\frac{1}{2}$). An antiderivative is $F(x) = \frac{1}{2}\ln|x - 1| - \frac{1}{2}\ln|x + 1| = \frac{1}{2}\ln\left|\frac{x - 1}{x + 1}\right|$. The poles $\pm 1$ lie outside $\left[-\frac{1}{2}, \frac{1}{2}\right]$, so the integrand is continuous there and the Fundamental Theorem applies. $F\left(\frac{1}{2}\right) = \frac{1}{2}\ln\left|\frac{-1/2}{3/2}\right| = \frac{1}{2}\ln\frac{1}{3}$ and $F\left(-\frac{1}{2}\right) = \frac{1}{2}\ln\left|\frac{-3/2}{1/2}\right| = \frac{1}{2}\ln 3$. So the integral is $\frac{1}{2}\ln\frac{1}{3} - \frac{1}{2}\ln 3 = -\ln 3$. The figure agrees: on this interval $x^2 - 1 < 0$, the curve is below the axis (it passes through $(0, -1)$), and a region below the axis contributes a NEGATIVE signed area; $-\ln 3 \approx -1.10$ against a width of $1$ and depths between $1$ and $\frac{4}{3}$ is consistent. Without absolute values, $\frac{x - 1}{x + 1}$ is negative on the whole interval and $\ln$ of it is undefined: the absolute values are not decoration, they are what makes the antiderivative exist on this interval.

d) $\frac{1}{(-x)^2 - 1} = \frac{1}{x^2 - 1}$: the integrand is even, so $\int_{-1/2}^{1/2} = 2\int_0^{1/2}$. With the same $F$: $2\left[F\left(\frac{1}{2}\right) - F(0)\right] = 2\left[\frac{1}{2}\ln\frac{1}{3} - \frac{1}{2}\ln 1\right] = \ln\frac{1}{3} = -\ln 3$. Same value, one evaluation fewer, and a useful cross-check of c).

a) Put the right side over the common denominator: $\frac{A}{x} + \frac{C}{x + 2} = \frac{A(x + 2) + Cx}{x(x + 2)} = \frac{x\left[A(x + 2) + Cx\right]}{x^2(x + 2)}$. The identity would therefore require $3x^2 - 4 = x\left[A(x + 2) + Cx\right]$ for every $x$. At $x = 0$ the right side is $0$ and the left side is $-4$: impossible, whatever $A$ and $C$ are. The deeper reason: every term of the proposed form has at most ONE factor $x$ in its denominator, so its sum cannot blow up like $\frac{1}{x^2}$ near $0$, while $\frac{3x^2 - 4}{x^2(x + 2)}$ behaves like $\frac{-4}{2x^2} = -\frac{2}{x^2}$ there. The factor $x^2$ is a REPEATED linear factor, and it opens two boxes.

b) Form: $\frac{3x^2 - 4}{x^2(x + 2)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x + 2}$, three unknowns for a denominator of degree $3$. Clearing denominators: $3x^2 - 4 = Ax(x + 2) + B(x + 2) + Cx^2$. At $x = 0$: $-4 = 2B$, so $B = -2$. At $x = -2$: $12 - 4 = 8 = 4C$, so $C = 2$. There is no third root to plug in, so compare the coefficients of $x^2$: $3 = A + C$, so $A = 1$. Check at $x = 1$: left $-1$; right $A \cdot 3 + B \cdot 3 + C = 3 - 6 + 2 = -1$. Now integrate each term by its own rule: $\int \frac{dx}{x} = \ln|x|$, $\int -\frac{2}{x^2}\,dx = \frac{2}{x}$, $\int \frac{2}{x + 2}\,dx = 2\ln|x + 2|$. So $\int \frac{3x^2 - 4}{x^3 + 2x^2}\,dx = \ln|x| + \frac{2}{x} + 2\ln|x + 2| + K$.

c) The poles $0$ and $-2$ are outside $[1, 2]$, so with $F(x) = \ln x + \frac{2}{x} + 2\ln(x + 2)$ on this interval: $F(2) = \ln 2 + 1 + 2\ln 4 = 5\ln 2 + 1$ and $F(1) = 0 + 2 + 2\ln 3$. Hence $\int_1^2 \frac{3x^2 - 4}{x^3 + 2x^2}\,dx = 5\ln 2 + 1 - 2 - 2\ln 3 = 5\ln 2 - 2\ln 3 - 1 = \ln\frac{32}{9} - 1$. Order of magnitude with $\ln 2 \approx 0.69$ and $\ln 3 \approx 1.10$: $3.47 - 2.20 - 1 \approx 0.27$. The integrand goes from $-\frac{1}{3}$ at $x = 1$ to $\frac{1}{2}$ at $x = 2$, so a small positive net area is what we should expect.

d) Proper (degree $2$ over $3$). Form: $\frac{x^2 + 1}{x(x - 1)^2} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{(x - 1)^2}$, so $x^2 + 1 = A(x - 1)^2 + Bx(x - 1) + Cx$. At $x = 0$: $1 = A$. At $x = 1$: $2 = C$. Coefficient of $x^2$: $1 = A + B$, so $B = 0$. The box $\frac{B}{x - 1}$ comes out empty for THIS numerator, but the form had to contain it: it is the numerator that decides the values, the denominator that decides the form, and a form without the $B$ term would have been marked wrong even though $B = 0$. So $\int \frac{x^2 + 1}{x(x - 1)^2}\,dx = \int \left(\frac{1}{x} + \frac{2}{(x - 1)^2}\right)dx = \ln|x| - \frac{2}{x - 1} + K$. The classmate's $2\ln|(x - 1)^2| = 4\ln|x - 1|$ has derivative $\frac{4}{x - 1}$, not $\frac{2}{(x - 1)^2}$: a squared factor integrates by the POWER rule, $\int (x - 1)^{-2}\,dx = -(x - 1)^{-1}$.

a) Proper. $x^3 + 9x = x(x^2 + 9)$, and $x^2 + 9$ has discriminant $-36 < 0$: irreducible. Form: $\frac{3x^2 + 2x + 9}{x(x^2 + 9)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 9}$, three unknowns for degree $3$. Clearing: $3x^2 + 2x + 9 = A(x^2 + 9) + (Bx + C)x$. At $x = 0$: $9 = 9A$, $A = 1$. Coefficient of $x^2$: $3 = A + B$, $B = 2$. Coefficient of $x$: $2 = C$. So the integrand is $\frac{1}{x} + \frac{2x}{x^2 + 9} + \frac{2}{x^2 + 9}$. The middle term is $\frac{u'}{u}$ with $u = x^2 + 9$, and the last one uses $a = 3$: $\int \frac{2}{x^2 + 9}\,dx = \frac{2}{3}\arctan\frac{x}{3}$, not $2\arctan\frac{x}{3}$ and not $\frac{2}{9}\arctan x$. Result: $\ln|x| + \ln(x^2 + 9) + \frac{2}{3}\arctan\frac{x}{3} + K$.

b) The discriminant of $x^2 + 2x + 5$ is $4 - 20 = -16 < 0$, so no partial fractions are possible: the quadratic IS the term. Complete the square: $x^2 + 2x + 5 = (x + 1)^2 + 4$. Split the numerator around the derivative $2x + 2$ of the denominator: $2x + 7 = (2x + 2) + 5$. Then $\int \frac{2x + 2}{x^2 + 2x + 5}\,dx = \ln(x^2 + 2x + 5)$, and with $u = x + 1$, $\int \frac{5}{(x + 1)^2 + 4}\,dx = \frac{5}{2}\arctan\frac{x + 1}{2}$. So $\int \frac{2x + 7}{x^2 + 2x + 5}\,dx = \ln(x^2 + 2x + 5) + \frac{5}{2}\arctan\frac{x + 1}{2} + K$. No absolute value is needed because $(x + 1)^2 + 4 \ge 4 > 0$ for every $x$: writing one is not wrong, but explaining why it is unnecessary shows the square was understood.

c) With $F(x) = \ln(x^2 + 2x + 5) + \frac{5}{2}\arctan\frac{x + 1}{2}$: $F(1) = \ln 8 + \frac{5}{2}\arctan 1 = \ln 8 + \frac{5\pi}{8}$ and $F(-1) = \ln 4 + \frac{5}{2}\arctan 0 = \ln 4$. So the area is $\ln 8 - \ln 4 + \frac{5\pi}{8} = \ln 2 + \frac{5\pi}{8}$. The bounds were chosen so that $x + 1$ runs from $0$ to $2$ and the arctangent lands on $\arctan 1 = \frac{\pi}{4}$. Check against the figure with the comparison property: on $[-1, 1]$ the curve stays between its value $\frac{9}{8}$ at $x = 1$ and a peak a little under $1.43$, so the area lies between $2 \times 1.125 = 2.25$ and $2 \times 1.43 \approx 2.86$. With $\ln 2 \approx 0.69$ and $\pi \approx 3.14$, $\ln 2 + \frac{5\pi}{8} \approx 0.69 + 1.96 = 2.65$: inside the bracket.

d) Here the discriminant is $4 + 12 = 16 > 0$: $x^2 + 2x - 3 = (x + 3)(x - 1)$ splits, and there is NO arctangent. Form $\frac{A}{x + 3} + \frac{B}{x - 1}$ with $2x + 7 = A(x - 1) + B(x + 3)$. At $x = -3$: $1 = -4A$, $A = -\frac{1}{4}$. At $x = 1$: $9 = 4B$, $B = \frac{9}{4}$. So $\int \frac{2x + 7}{x^2 + 2x - 3}\,dx = -\frac{1}{4}\ln|x + 3| + \frac{9}{4}\ln|x - 1| + K$. Two denominators that differ only in their constant term give answers of a different NATURE: the discriminant, computed before anything else, decides between logarithms and an arctangent. Completing the square on $x^2 + 2x - 3 = (x + 1)^2 - 4$ would lead to a DIFFERENCE of squares, which is the signal that the quadratic factors.

a) Let $u = \sqrt{x}$, so $x = u^2$, $dx = 2u\,du$, and the bounds become $u = 1$ and $u = 2$. The integral becomes $\int_1^2 \frac{2u\,du}{u^2 + 3u + 2}$, a proper rational function of $u$. Factor: $u^2 + 3u + 2 = (u + 1)(u + 2)$. Form: $\frac{2u}{(u + 1)(u + 2)} = \frac{A}{u + 1} + \frac{B}{u + 2}$, $2u = A(u + 2) + B(u + 1)$. At $u = -1$: $-2 = A$. At $u = -2$: $-4 = -B$, $B = 4$. So the integral is $\left[-2\ln(u + 1) + 4\ln(u + 2)\right]_1^2 = (-2\ln 3 + 4\ln 4) - (-2\ln 2 + 4\ln 3) = 10\ln 2 - 6\ln 3 = \ln\frac{2^{10}}{3^6} = \ln\frac{1024}{729}$. It is positive since $1024 > 729$, and small: the integrand lies between $\frac{1}{12}$ (at $x = 4$) and $\frac{1}{6}$ (at $x = 1$) on an interval of length $3$, so the value is between $0.25$ and $0.5$, and $\ln\frac{1024}{729} \approx 10(0.69) - 6(1.10) = 0.3$. The trap is the factor $2u$ from $dx$: without it the integrand in $u$ would be $\frac{1}{(u + 1)(u + 2)}$ and every constant would be wrong.

b) $u = \sqrt{x}$, $x = u^2$, $dx = 2u\,du$: $\int \frac{u \cdot 2u}{u^2 - 4}\,du = \int \frac{2u^2}{u^2 - 4}\,du$. The substitution produced an IMPROPER fraction (degree $2$ over degree $2$), so the degree check comes back first: $2u^2 = 2(u^2 - 4) + 8$, hence $\frac{2u^2}{u^2 - 4} = 2 + \frac{8}{(u - 2)(u + 2)}$. Cover-up: at $u = 2$, $\frac{8}{4} = 2$; at $u = -2$, $\frac{8}{-4} = -2$. So the integral is $\int \left(2 + \frac{2}{u - 2} - \frac{2}{u + 2}\right)du = 2u + 2\ln|u - 2| - 2\ln|u + 2| + K$. Back to $x$, and since $\sqrt{x} + 2 > 0$: $\int \frac{\sqrt{x}}{x - 4}\,dx = 2\sqrt{x} + 2\ln\left|\frac{\sqrt{x} - 2}{\sqrt{x} + 2}\right| + K$. The absolute value on $\sqrt{x} - 2$ stays: it is negative for $0 < x < 4$.