Integration by parts and partial fractions

In week 12, substitution was the reverse of the chain rule. This week there are two new techniques. Integration by parts is the reverse of the product rule; it solves integrals like $x\,e^x$, $x\sin x$, $\ln x$. Partial fractions breaks a rational function into pieces that are easy to integrate. The final has long but mechanical questions on both; whoever chooses the right start gets the result:

1. $\int u\,dv=uv-\int v\,du$. Choose $u$ in LIATE order.
2. The integral left after parts should be easier. If it got harder, you chose $u$ the wrong way round.
3. For a rational function, first look at the degrees: if the numerator is larger or equal, divide first.
4. The type of the denominator’s factors determines the shape of the fractions.

Integration by parts

Product rule: $(uv)'=u'v+uv'$. Integrate both sides and rearrange:

$\int u\,dv=uv-\int v\,du$

For a definite integral:

$\int_a^bu\,dv=\Big[uv\Big]_a^b-\int_a^bv\,du$

The idea: split the integrand into two parts. Let $u$ be the part that simplifies when differentiated, and $dv$ the part that can be integrated.

Choosing $u$: LIATE

The one that comes first in the list becomes $u$:

Example 1. $\displaystyle\int x\sin x\,dx$. A comes before T: $u=x$, $dv=\sin x\,dx$.

u = x          dv = sin x dx
du = dx        v = -cos x
∫ x sin x dx = -x cos x - ∫ (-cos x) dx = -x cos x + sin x + C

If you had chosen the other way round ($u=\sin x$, $dv=x\,dx$), the remaining integral would be $\int\tfrac{x^2}{2}\cos x\,dx$, which is harder.

Example 2. $\displaystyle\int x^2\ln x\,dx$. L comes before A: $u=\ln x$, $dv=x^2dx$, $v=\tfrac{x^3}{3}$.

$\int x^2\ln x\,dx=\frac{x^3}{3}\ln x-\int\frac{x^3}{3}\cdot\frac1x\,dx=\frac{x^3\ln x}{3}-\frac{x^3}{9}+C$

$\ln x$ is hard to integrate and easy to differentiate. That is why the logarithm is always $u$.

The tabular method

For $x^n$ times $e^{ax}$, $\sin ax$ or $\cos ax$, you would have to do integration by parts $n$ times. The table speeds this up: in the left column differentiate $u$ until it reaches zero, in the right column integrate $dv$ repeatedly, multiply diagonally, and alternate the signs $+,-,+,-$.

Example 3. $\displaystyle\int x^3e^x\,dx$.

derivative   integral     sign
x^3          e^x
3x^2         e^x            +   ->  x^3 e^x
6x           e^x            -   ->  3x^2 e^x
6            e^x            +   ->  6x e^x
0            e^x            -   ->  6 e^x
result: e^x (x^3 - 3x^2 + 6x - 6) + C

Cyclic integrals

For $e^{ax}\sin bx$ or $e^{ax}\cos bx$, after two integrations by parts the integral you started with comes back. Then call the integral $I$ and solve the equation.

Example 4. $I=\displaystyle\int e^{2x}\cos x\,dx$.

First step: $u=e^{2x}$, $dv=\cos x\,dx$. $I=e^{2x}\sin x-2\int e^{2x}\sin x\,dx$.

Second step (same choice logic, $u=e^{2x}$): $\int e^{2x}\sin x\,dx=-e^{2x}\cos x+2I$.

Combine: $I=e^{2x}\sin x-2(-e^{2x}\cos x+2I)=e^{2x}\sin x+2e^{2x}\cos x-4I$. So $5I=e^{2x}(\sin x+2\cos x)$:

$I=\frac{e^{2x}(2\cos x+\sin x)}{5}+C$

PITFALL · changing $u$ in the second step. In a cyclic integral, if you chose $u=e^{2x}$ in the first step, make the exponential part $u$ in the second step too. If you switch, you undo the first step and get $I=I$.

Example 5 (definite). $\displaystyle\int_0^1xe^x\,dx=\Big[xe^x\Big]_0^1-\int_0^1e^x\,dx=e-(e-1)=1$.

Partial fractions

Of the form $\dfrac{P(x)}{Q(x)}$, with $P$ and $Q$ polynomials. Strategy: write the fraction as a sum of simple fractions whose denominators are the factors of $Q$. The integral of each piece gives $\ln$ or $\arctan$.

Step 0: check the degrees

If the numerator’s degree is larger than or equal to the denominator’s, first do polynomial division: $\dfrac PQ=(\text{quotient})+\dfrac{\text{remainder}}{Q}$. The decomposition is applied only to the remainder fraction.

Factor types

To find the coefficients, multiply both sides by the denominator. Then plug in the roots of the denominator (each root gives one coefficient directly), and for the remaining coefficients equate coefficients such as the $x^2$ and constant terms.

Example 6 (distinct linear factors). $\displaystyle\int\frac{3x-1}{x^2-x-2}\,dx$.

x^2 - x - 2 = (x - 2)(x + 1)
(3x - 1) = A(x + 1) + B(x - 2)
x = 2:   5 = 3A   ->  A = 5/3
x = -1: -4 = -3B  ->  B = 4/3
∫ = (5/3) ln|x - 2| + (4/3) ln|x + 1| + C

Example 7 (repeated factor). $\displaystyle\int\frac{2x+3}{(x+1)^2}\,dx$.

$2x+3=A(x+1)+B$. $x=-1$: $B=1$. Coefficient of $x$: $A=2$. So $\dfrac{2}{x+1}+\dfrac{1}{(x+1)^2}$, and the integral is $2\ln|x+1|-\dfrac{1}{x+1}+C$.

Example 8 (irreducible quadratic). $\displaystyle\int\frac{x^2+2x+3}{x(x^2+1)}\,dx$.

$x^2+2x+3=A(x^2+1)+(Bx+C)x$. $x=0$: $A=3$. $x^2$: $1=A+B \Rightarrow B=-2$. $x$: $2=C$. The fractions are $\dfrac3x+\dfrac{-2x+2}{x^2+1}$:

$\int=3\ln|x|-\ln(x^2+1)+2\arctan x+C$

The part $\dfrac{-2x}{x^2+1}$ gave $-\ln(x^2+1)$ by substitution ($u=x^2+1$), and the part $\dfrac{2}{x^2+1}$ gave $2\arctan x$.

PITFALL · writing only a constant in the numerator of an irreducible factor. If you write $\dfrac{C}{x^2+1}$, the equation won’t match. The numerator of a quadratic factor is always $Bx+C$.

TIP · complete the square in the denominator. For an irreducible $x^2+bx+c$ write $\left(x+\tfrac b2\right)^2+k^2$. $\displaystyle\int\frac{dx}{u^2+k^2}=\frac1k\arctan\frac uk+C$.

Common mistakes