Integration by parts and partial fractions Knowledge-Based Study

Study notes

1. Integration by parts is an analytical technique derived directly from the product rule for differentiation. For two differentiable functions $u(x)$ and $v(x)$, the product rule states:

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

   Integrating both sides with respect to $x$ yields:

   $$u(x)v(x) = \int v(x)u'(x)\,dx + \int u(x)v'(x)\,dx$$

   Rearranging this identity gives the standard integration by parts formula:

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

   This method transforms an integral $\int u\,dv$ into an expression containing a new integral $\int v\,du$, which is often simpler to evaluate.

   A critical step in applying integration by parts is choosing which factor serves as $u$ and which serves as $dv$. A widely used heuristic for selecting $u$ is the LIATE rule, ordered by priority:

   • L: Logarithmic functions (e.g., $\ln(x)$)
   • I: Inverse trigonometric functions (e.g., $\arctan(x)$, $\arcsin(x)$)
   • A: Algebraic/polynomial functions (e.g., $x^2$, $3x$)
   • T: Trigonometric functions (e.g., $\sin(x)$, $\cos(x)$)
   • E: Exponential functions (e.g., $e^x$, $2^x$)

   The remaining factor, along with $dx$, is assigned to $dv$, which must be straightforward to integrate.

   Repeated Integration and the Tabular Method

   When $u$ is a polynomial and $dv$ can be repeatedly integrated without terminating (such as $\int x^3 e^{2x}\,dx$), the tabular method streamlines repeated applications. One column lists successive derivatives of $u$ down to 0, while a second column lists successive antiderivatives of $dv$. Products along alternating-sign diagonals yield the complete antiderivative.

   Integration by partial fractions is used to evaluate integrals of rational functions $\frac{P(x)}{Q(x)}$, where $P(x)$ and $Q(x)$ are polynomials.

   Step 1: Check Degree

   If $\deg(P) \ge \deg(Q)$, perform polynomial long division first to express the integrand as a polynomial plus a proper rational expression.

   Step 2: Distinct Linear Factors

   If the denominator $Q(x)$ factors completely into distinct linear factors $(a_1x + b_1)(a_2x + b_2)\cdots(a_kx + b_k)$, decompose the fraction as:

   $$\frac{P(x)}{Q(x)} = \frac{A_1}{a_1x + b_1} + \frac{A_2}{a_2x + b_2} + \cdots + \frac{A_k}{a_kx + b_k}$$

   The constants $A_i$ are determined either by clearing denominators and equating coefficients or by substituting convenient roots of the linear factors (the Heaviside cover-up method). Once decomposed, each term integrates directly into logarithmic form:

   $$\int \frac{A}{ax + b}\,dx = \frac{A}{a}\ln|ax + b| + C$$

   When the factored denominator contains higher powers or quadratics that cannot be factored over real numbers, the setup changes accordingly:

   Repeated Linear Factors

   For each factor of the form $(ax + b)^m$, include $m$ distinct terms with increasing powers in the denominator:

   $$\frac{A_1}{ax + b} + \frac{A_2}{(ax + b)^2} + \cdots + \frac{A_m}{(ax + b)^m}$$

   Higher-power terms integrate using the power rule rather than logarithms.

   Irreducible Quadratic Factors

   For each distinct irreducible quadratic factor $ax^2 + bx + c$ (where $b^2 - 4ac < 0$), assign a linear numerator:

   $$\frac{Ax + B}{ax^2 + bx + c}$$

   These integrals typically split into two parts: one evaluated via natural substitution and the other via completing the square leading to an $\arctan$ form:

   $$\int \frac{1}{x^2 + a^2}\,dx = \frac{1}{a}\arctan\left(\frac{x}{a}\right) + C$$

   Proficiency in Calculus BC requires choosing efficiently between integration techniques:

   • Use Integration by Parts when the integrand is a product of dissimilar functions (e.g., algebraic times trigonometric, exponential times logarithmic), or when integrating isolated transcendental functions such as $\int \ln(x)\,dx$ or $\int \arctan(x)\,dx$ by setting $dv = dx$.
   • Watch for looping integrals (e.g., $\int e^x \cos(x)\,dx$), where applying integration by parts twice returns a multiple of the original integral, allowing algebraic solving for the unknown integral.
   • Use Partial Fractions strictly for rational functions after confirming that simpler substitutions (such as $u = Q(x)$) do not apply.