7.2 Integration by Parts & Partial Fractions

Integration by Parts & Partial Fractions

When an integrand consists of a product of dissimilar functional families (such as algebraic polynomials multiplied by exponentials, logarithms, or circular trigonometric functions), or rational fractions that cannot be reduced by simple substitutions, standard elementary anti-differentiation fails. On the Pakistan ECAT, mastering advanced product handling and algebraic decomposition is vital. This section details integration by parts, the ILATE priority rule, the rapid tabular DI method, the master exponential theorem, and systematic partial fraction decomposition utilizing the Heaviside cover-up method.

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1. Integration by Parts: Derivation & Mechanics

Analytical Derivation from the Product Rule

Integration by parts is the integral calculus analog of the Leibniz product rule for differentiation. Let $u(x)$ and $v(x)$ be two continuously differentiable functions of $x$. Differentiating their product yields:

$\frac{d}{dx}[u(x) v(x)] = u(x) \frac{dv}{dx} + v(x) \frac{du}{dx}$

Integrating both sides with respect to $x$:

$u(x) v(x) = \int u(x) \frac{dv}{dx} dx + \int v(x) \frac{du}{dx} dx$

Rearranging to isolate the first integral produces the classical integration by parts formula:

$\int u \frac{dv}{dx} dx = u v - \int v \frac{du}{dx} dx$

Or in compact differential notation:

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

Verbal Formulation

In traditional intermediate textbooks, this rule is expressed as: $\int (\text{1st function}) \cdot (\text{2nd function}) dx = (\text{1st}) \cdot \int (\text{2nd}) dx - \int \left[ \frac{d}{dx}(\text{1st}) \cdot \int (\text{2nd}) dx \right] dx$

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2. The ILATE Priority Hierarchy

The strategic success of integration by parts hinges entirely upon the choice of $u$ (the 'first function' to differentiate) versus $dv$ (the 'second function' to integrate). If chosen poorly, the resulting integral $\int v du$ becomes significantly more complex than the original problem.

To ensure systematic selection, mathematicians established the ILATE priority hierarchy. The function appearing highest in the following order must be chosen as $u(x)$:

1. I — Inverse Trigonometric Functions: $\arcsin x, \arccos x, \arctan x, \text{arcsec } x$
2. L — Logarithmic Functions: $\ln x, \log_a x$
3. A — Algebraic Functions: $x^n, ax + b, x^2 + 5, \sqrt{x}$
4. T — Trigonometric Functions: $\sin x, \cos x, \tan x, \sec x$
5. E — Exponential Functions: $e^x, a^x, e^{kx}$

Why ILATE Works

Functions at the top of the list (Inverse trigonometric and Logarithmic) have no elementary integrals of their own, but differentiate into simple algebraic fractions. Conversely, functions at the bottom (Exponential and Trigonometric) can be integrated repeatedly without increasing their algebraic complexity.

Integrating Single Transcendental Functions via the Factor of Unity ($1$)

A classical ECAT technique applies integration by parts to single inverse trigonometric or logarithmic functions by introducing the phantom algebraic factor $1 = x^0$ as the second function ($dv = 1 dx \implies v = x$):

1. $\int \ln x dx$:

   • Choose $u = \ln x$ and $dv = 1 dx \implies du = \frac{1}{x} dx, v = x$.
   • $\int \ln x dx = x \ln x - \int x \left(\frac{1}{x}\right) dx = x \ln x - \int 1 dx = x \ln x - x + c = x(\ln x - 1) + c$.

2. $\int \arctan x dx$:

   • Choose $u = \arctan x$ and $dv = 1 dx \implies du = \frac{1}{1 + x^2} dx, v = x$.
   • $\int \arctan x dx = x \arctan x - \int \frac{x}{1 + x^2} dx = x \arctan x - \frac{1}{2}\ln(1 + x^2) + c$.

3. $\int \arcsin x dx$:

   • Choose $u = \arcsin x$ and $dv = 1 dx \implies du = \frac{1}{\sqrt{1 - x^2}} dx, v = x$.
   • $\int \arcsin x dx = x \arcsin x - \int \frac{x}{\sqrt{1 - x^2}} dx = x \arcsin x + \sqrt{1 - x^2} + c$.

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3. Tabular Integration (The DI Method)

When integrating products of polynomials $P(x)$ and functions that can be integrated indefinitely without termination (such as $e^{ax}$, $\sin(bx)$, or $\cos(bx)$), applying standard integration by parts repeatedly generates nested minus signs and high rates of algebraic error. The Tabular Method (or DI Method) resolves this in seconds.

Algorithmic Procedure

1. Construct two columns labeled D (Differentiate) and I (Integrate).
2. Place the polynomial $P(x)$ in the D column and repeatedly differentiate until reaching $0$.
3. Place the transcendental factor ($e^{ax}, \sin bx, \cos bx$) in the I column and repeatedly integrate an equal number of times.
4. Assign an alternating sign column starting with $+$, followed by $-, +, -, \dots$.
5. Connect each entry in row $k$ of column D with the entry in row $k+1$ of column I with a diagonal arrow.
6. Multiply terms along each diagonal arrow, apply the indicated row sign, and sum the results.

Comprehensive Walkthrough: Evaluate $\int x^3 e^{2x} dx$

Summing all products and factoring out $\frac{e^{2x}}{8}$:

$\int x^3 e^{2x} dx = e^{2x} \left( \frac{x^3}{2} - \frac{3x^2}{4} + \frac{3x}{4} - \frac{3}{8} \right) + c = \frac{e^{2x}}{8}(4x^3 - 6x^2 + 6x - 3) + c$

The table replaces three successive rounds of integration by parts with one compact layout, which removes most sign errors.

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4. Cyclic Integrals

When neither factor differentiates to zero—specifically when integrating products of exponential and trigonometric functions ($e^{ax}\sin bx$ or $e^{ax}\cos bx$)—applying integration by parts twice recreates a scalar multiple of the original integral. This is known as a cyclic integral.

Closed-Form ECAT Formulas

Candidates should memorize these closed forms directly:

$\int e^{ax}\sin(bx) dx = \frac{e^{ax}}{a^2 + b^2} [a\sin(bx) - b\cos(bx)] + c$

$\int e^{ax}\cos(bx) dx = \frac{e^{ax}}{a^2 + b^2} [a\cos(bx) + b\sin(bx)] + c$

Memory Aid: The denominator is always the sum of squares $a^2 + b^2$. The bracketed term begins with $a$ multiplying the original trigonometric function, followed by $-b$ times the derivative of sine (which is $-b\cos bx$) or $+b$ times the derivative of cosine (reversing sign, yielding $+b\sin bx$).

The Integral of $\sec^3 x$

Another famous cyclic integral is $\int \sec^3 x dx$: $\int \sec^3 x dx = \int \sec x \cdot \sec^2 x dx$ Let $u = \sec x, dv = \sec^2 x dx \implies du = \sec x \tan x dx, v = \tan x$: $\int \sec^3 x dx = \sec x \tan x - \int \sec x \tan^2 x dx = \sec x \tan x - \int \sec x (\sec^2 x - 1) dx$ $\int \sec^3 x dx = \sec x \tan x - \int \sec^3 x dx + \int \sec x dx$ Adding $\int \sec^3 x dx$ to both sides: $2\int \sec^3 x dx = \sec x \tan x + \ln|\sec x + \tan x|$

$\int \sec^3 x dx = \frac{1}{2}\sec x \tan x + \frac{1}{2}\ln|\sec x + \tan x| + c$

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5. The Master Exponential Theorems

A very useful integration shortcut is the exponential product identity, which follows directly from the product rule.

Master Theorem 1: Base Exponential Form

$\int e^x [f(x) + f'(x)] dx = e^x f(x) + c$

Proof: Expand into two separate integrals: $\int e^x f(x) dx + \int e^x f'(x) dx$. Apply integration by parts to the first integral, setting $u = f(x)$ and $dv = e^x dx$: $\int e^x f(x) dx = f(x) e^x - \int f'(x) e^x dx$ Substituting this back cancels the second integral identically: $\left[ e^x f(x) - \int e^x f'(x) dx \right] + \int e^x f'(x) dx = e^x f(x) + c$

Master Theorem 2: Generalized Scaling Form

When an arbitrary scaling constant $k$ multiplies $f(x)$:

$\int e^{kx} [k f(x) + f'(x)] dx = e^{kx} f(x) + c$

ECAT Classic Disguised Expressions

Examiners routinely conceal this theorem inside rational fractions:

1. $\int e^x \left(\frac{1}{x} - \frac{1}{x^2}\right) dx$: Here $f(x) = \frac{1}{x}$ and $f'(x) = -\frac{1}{x^2}$. Answer: $\frac{e^x}{x} + c$.
2. $\int e^x \left(\frac{x}{(x+1)^2}\right) dx$: Split the numerator $x = (x + 1) - 1$: $\frac{x}{(x+1)^2} = \frac{x+1}{(x+1)^2} - \frac{1}{(x+1)^2} = \frac{1}{x+1} - \frac{1}{(x+1)^2}$ Here $f(x) = \frac{1}{x+1} \implies f'(x) = -\frac{1}{(x+1)^2}$. Answer: $\frac{e^x}{x+1} + c$.
3. $\int e^x (\tan x + \sec^2 x) dx$: Here $f(x) = \tan x, f'(x) = \sec^2 x$. Answer: $e^x \tan x + c$.
4. $\int e^x \left(\frac{1 + \sin x}{1 + \cos x}\right) dx$: Use half-angle identities: $\frac{1 + 2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)} = \frac{1}{2}\sec^2(x/2) + \tan(x/2)$ Here $f(x) = \tan(x/2)$, whose derivative is $f'(x) = \frac{1}{2}\sec^2(x/2)$. Answer: $e^x \tan(x/2) + c$.

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6. Partial Fraction Decomposition for Rational Functions

A rational function is an algebraic quotient $\frac{P(x)}{Q(x)}$ where $P(x)$ and $Q(x)$ are polynomials.

• Proper Rational Fraction: $\deg(P) < \deg(Q)$.
• Improper Rational Fraction: $\deg(P) \ge \deg(Q)$.

Mandatory Rule: If a rational fraction is improper (degree of numerator $\ge$ degree of denominator), polynomial long division is strictly required first before applying partial fractions: $\frac{P(x)}{Q(x)} = S(x) + \frac{R(x)}{Q(x)}$, where $\deg(R) < \deg(Q)$.

Classification of Proper Partial Fraction Decompositions

The Heaviside Cover-Up Method

For proper rational fractions with distinct, non-repeated linear factors, the classical method of equating undetermined coefficients wastes far too much time. Oliver Heaviside's cover-up method computes coefficients instantly in one step.

To find the numerator coefficient $A_i$ corresponding to the linear factor $(x - r_i)$:

1. Cover (remove) the factor $(x - r_i)$ from the denominator of the original fraction.
2. Evaluate the remainder of the expression at the root $x = r_i$.

$A_i = \lim_{x \to r_i} \left[ (x - r_i) \frac{P(x)}{Q(x)} \right]$

Example: Decompose $\frac{5x - 2}{(x - 2)(x + 1)}$.

• To find $A$ above $(x - 2)$: Cover $(x - 2)$ and set $x = 2$: $A = \left. \frac{5x - 2}{x + 1} \right|_{x=2} = \frac{5(2) - 2}{2 + 1} = \frac{8}{3}$
• To find $B$ above $(x + 1)$: Cover $(x + 1)$ and set $x = -1$: $B = \left. \frac{5x - 2}{x - 2} \right|_{x=-1} = \frac{5(-1) - 2}{-1 - 2} = \frac{-7}{-3} = \frac{7}{3}$
• Instant Decomposition: $\frac{5x - 2}{(x - 2)(x + 1)} = \frac{8/3}{x - 2} + \frac{7/3}{x + 1}$. Done in 10 seconds!

Repeated Linear Factor Example

Decompose and integrate $\frac{x + 3}{(x - 1)^2(x + 1)}$. Set

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

• Cover-up gives the top-power coefficient and the distinct factor: $B = \left.\frac{x + 3}{x + 1}\right|_{x=1} = 2$ and $C = \left.\frac{x + 3}{(x - 1)^2}\right|_{x=-1} = \frac{2}{4} = \frac{1}{2}$.
• The remaining coefficient comes from comparing $x^2$ terms in $x + 3 = A(x - 1)(x + 1) + B(x + 1) + C(x - 1)^2$: $0 = A + C$, so $A = -\frac{1}{2}$.
• Check at $x = 0$: $3 = -A + B + C = \frac{1}{2} + 2 + \frac{1}{2}$. Correct.

$\int \frac{x + 3}{(x - 1)^2(x + 1)}\,dx = -\frac{1}{2}\ln|x - 1| - \frac{2}{x - 1} + \frac{1}{2}\ln|x + 1| + c$

Irreducible Quadratic Factor Example

$\frac{1}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$

Cover-up gives $A = 1$. Then $1 = (x^2 + 1) + (Bx + C)x$ forces $B = -1$ and $C = 0$, so

$\int \frac{dx}{x(x^2 + 1)} = \ln|x| - \frac{1}{2}\ln(x^2 + 1) + c$