Integration by Parts

For continuously differentiable $u,v$ on $[a,b]$, integrate $(uv)'=u'v+uv'$ and rearrange:

$$\int_a^b u(x)v'(x)\,dx=[u(x)v(x)]_a^b-\int_a^b v(x)u'(x)\,dx.$$

Choose $u$ so differentiation simplifies it and $dv$ so it has an accessible antiderivative. The new integral should be easier; the rule is an identity, not a guarantee of progress. In the first example, differentiating $x$ removes the polynomial factor. In the logarithm example, write the integrand as $\ln x\cdot1$.

For the quadratic substitutions below take $a,b>0$ and choose an angle branch before removing absolute values. For sine and tangent substitutions use $-\pi/2<\theta<\pi/2$. For the secant substitution on $x>a/b$, use $0<\theta<\pi/2$; the negative-$x$ branch must be handled separately. In the worked sine example the real domain is $0<|x|<2/3$ and $\theta=\arcsin(3x/2)$ ensures $\cos\theta>0$. The cotangent relation also holds for negative $x$ by $\cot\theta=\cos\theta/\sin\theta$, not by assigning a negative side length to a triangle.

For partial fractions, divide polynomials first if the numerator degree is not smaller. Factor over the reals; quadratic factors in the table must be irreducible ($b^2-4ac<0$). Each repeated factor needs all powers up to its multiplicity. The worked rational example is valid separately on $x<1$ and $x>1$.

Integration by Parts

Integration by parts is a technique based on the product rule for differentiation. The formula is:

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

The choice of $u$ and $dv$ is crucial, and differentiating $u$ and integrating $dv$ gives us $du$ and $v$, respectively.

Example I

$$\int x e^{-x} \, dx$$

Let $u = x$ which implies $du = dx$. Choose $dv = e^{-x} dx$ then $v = -e^{-x}$.

By the integration by parts formula $\int u \, dv = uv - \int v \, du$, we get:

$$\int x e^{-x} \, dx = uv - \int v \, du$$ $$\int x e^{-x} \, dx = -x e^{-x} - \int -e^{-x} \, dx$$

Since $\int -e^{-x}\,dx=e^{-x}$, subtracting that integral gives:

$$\int x e^{-x} \, dx = -x e^{-x} - e^{-x} + C = -(x+1)e^{-x}+C$$

Where $C$ is the constant of integration.

Example II

$$\int_{3}^{5} \ln(x) \, dx$$

Let $u = \ln(x)$ which implies $du = \frac{1}{x}dx$. Choose $dv = dx$ then $v = x$.

Using the integration by parts formula $\int_{a}^{b} u \, dv = uv \bigg|_{a}^{b} - \int_{a}^{b} v \, du$, we obtain:

$$\int_{3}^{5} \ln(x) \, dx = x\ln(x) \bigg|_{3}^{5} - \int_{3}^{5} x \, \frac{1}{x} \, dx$$

Simplifying the integral $\int_{3}^{5} x \, \frac{1}{x} \, dx$ to $\int_{3}^{5} dx$, we have:

$$\int_{3}^{5} \ln(x) \, dx = x\ln(x) \bigg|_{3}^{5} - \int_{3}^{5} dx$$ $$\int_{3}^{5} \ln(x) \, dx = x\ln(x) \bigg|_{3}^{5} - x \bigg|_{3}^{5}$$

Subtracting $5 - 3$ from $5\ln(5) - 3\ln(3)$, the final result is:

$$\int_{3}^{5} \ln(x) \, dx = 5\ln(5) - 3\ln(3) - (5 - 3)$$ $$\int_{3}^{5} \ln(x) \, dx = 5\ln(5) - 3\ln(3) - 2$$

Trigonometric Substitutions in Integration

Substitutions Based on Quadratic Form

Case 1: $\sqrt{a^2 - b^2x^2}$

• Substitution: $x = \frac{a}{b} \sin(\theta)$
• Identity: $\cos^2(\theta) = 1 - \sin^2(\theta)$

Case 2: $\sqrt{b^2x^2 - a^2}$

• Substitution: $x = \frac{a}{b} \sec(\theta)$
• Identity: $\tan^2(\theta) = \sec^2(\theta) - 1$

Case 3: $\sqrt{a^2 + b^2x^2}$

• Substitution: $x = \frac{a}{b} \tan(\theta)$
• Identity: $\sec^2(\theta) = 1 + \tan^2(\theta)$

Example

$$\int \frac{16}{x^2 \sqrt{4-9x^2}} \, dx$$

Let $x = \frac{2}{3} \sin(\theta)$ which implies $dx = \frac{2}{3}\cos(\theta)d\theta$.

$$\sqrt{4-9x^2}=\sqrt{4-4\sin^2(\theta)}=\sqrt{4\cos^2(\theta)}= 2 \mid \cos(\theta) \mid$$

In this case we have $\sqrt{4-9x^2} = 2\cos(\theta)$

$$\int \frac{16}{x^2 \sqrt{4-9x^2}} \, dx = \int \frac{16}{\left(\frac{2}{3}\sin(\theta)\right)^2 \cdot 2\cos(\theta)} \cdot \frac{2}{3}\cos(\theta) \, d\theta = \int \frac{12}{\sin^2(\theta)} \, d\theta$$ $$\int \frac{16}{\left(\frac{2}{3}\sin(\theta)\right)^2 \cdot 2\cos(\theta)} \cdot \frac{2}{3}\cos(\theta) \, d\theta = \int 12 \csc^2(\theta) \, d\theta$$

This simplification uses the identity $\sin^2(\theta) + \cos^2(\theta) = 1$ and the definition of $\csc(\theta) = \frac{1}{\sin(\theta)}$.

The antiderivative of $\csc^2(\theta)$ is $- \cot(\theta)$. So, the evaluation is:

$$\int 12 \csc^2(\theta) \, d\theta = -12 \cot(\theta) + C$$

Using Right Triangle Trigonometry

To convert back to $x$, we use the right triangle relation where $\sin(\theta) = \frac{3x}{2}$ from the original substitution.

The triangle formed by the substitution suggests that:

$$\cot(\theta) = \frac{\sqrt{4 - 9x^2}}{3x}$$

Final Integral in terms of X

Substituting $\cot(\theta)$ back into the integral expression:

$$-12 \cot(\theta) + C = -4\sqrt{4 - 9x^2} \frac{1}{x} + C$$

Thus, the final antiderivative in terms of $x$ is:

$$\int \frac{16}{x^2 \sqrt{4 - 9x^2}} \, dx = -4 \frac{\sqrt{4 - 9x^2}}{x} + C$$

Open full-size image

Match the labels to the worked substitution: replace the figure’s a by 2 and its x by 3x. For 0 < x < 2/3, the adjacent side is √(4 − 9x²), so cot θ is adjacent divided by opposite. For negative x, use the signed trigonometric identity already stated above rather than negative side lengths. Access for free at OpenStax.

Gilbert Strang & Edwin Herman, OpenStax, Calculus Volume 2, Figure 3.4. · CC BY-NC-SA 4.0

Partial Fraction Decomposition for Integration

Concept

Partial fractions is a technique used to simplify the integration of rational expressions, $\frac{P(x)}{Q(x)}$, where the degree of $P(x)$ is less than the degree of $Q(x)$.

Steps for Decomposition

1. Factor the denominator $Q(x)$ as completely as possible.
2. Write down a partial fraction for each factor in $Q(x)$ using constants to represent the numerators.
3. Solve for the constants by clearing the denominators and equating coefficients for corresponding powers of $x$.

Table for Decomposition

Factor of $Q(x)$	Term in Partial Fraction Decomposition (P.F.D)
$ax + b$	$\frac{A}{ax + b}$
$ax^2 + bx + c$	$\frac{Ax + B}{ax^2 + bx + c}$
$(ax + b)^k$	$\frac{A_1}{ax + b} + \frac{A_2}{(ax + b)^2} + \cdots + \frac{A_k}{(ax + b)^k}$
$(ax^2 + bx + c)^k$	$\frac{A_1x + B_1}{ax^2 + bx + c} + \cdots + \frac{A_kx + B_k}{(ax^2 + bx + c)^k}$

Example of Integration Using Partial Fraction

$$\int \frac{7x^2 + 13x}{(x - 1)(x^2 + 4)} \, dx$$

Partial Fraction Decomposition

To decompose the function, assume it can be written as the sum of fractions:

$$\frac{7x^2 + 13x}{(x - 1)(x^2 + 4)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 4}$$

Where $A$, $B$, and $C$ are constants to be determined.

Finding Constants

Multiply both sides by the common denominator $(x - 1)(x^2 + 4)$ and equate the numerators:

$$7x^2 + 13x = A(x^2 + 4) + (Bx + C)(x - 1)$$

Expanding the right side and collecting like terms gives:

$$7x^2 + 13x = (A + B)x^2 + (C - B)x + 4A - C$$

Setting Coefficients Equal

Match the coefficients of corresponding powers of $x$ from both sides of the equation:

$$\begin{align*} x^2: & \quad A + B = 7 \\ x: & \quad C - B = 13 \\ \text{Constant}: & \quad 4A - C = 0 \end{align*}$$

Solving the System of Equations

Solve this system of linear equations to find the values of $A$, $B$, and $C$:

$$\begin{align*} A & = 4 \\ B & = 3 \\ C & = 16 \end{align*}$$

Rewriting the Integral

Substitute the values of $A$, $B$, and $C$ back into the partial fractions:

$$\int \frac{4}{x - 1} + \frac{3x + 16}{x^2 + 4} \, dx$$

Integrating Each Term

1. For $\frac{4}{x - 1}$, the antiderivative is $4 \ln|x - 1|$.
2. For $\frac{3x}{x^2 + 4}$, use substitution $u = x^2 + 4$ to find the antiderivative $\frac{3}{2} \ln|x^2 + 4|$.
3. For $\frac{16}{x^2 + 4}$, recognize it as the derivative of $8\tan^{-1}\left(\frac{x}{2}\right)$ because $\frac{d}{dx}\left[\tan^{-1}\left(\frac{x}{2}\right)\right] = \frac{2}{x^2 + 4}$.

Final Answer

Combine all antiderivatives and add the constant of integration to get the final solution:

$$\int \frac{7x^2 + 13x}{(x - 1)(x^2 + 4)} \, dx = 4 \ln|x - 1| + \frac{3}{2} \ln|x^2 + 4| + 8\tan^{-1}\left(\frac{x}{2}\right) + C$$

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