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Advanced Integration Techniques

Combination of Substitution and Integration by Parts, Recursive and Cancellation Techniques

1. Combination of Substitution and Integration by Parts

Definition 1.1: Combination Strategy

When the integrand contains both complex structure and product form, we typically need to first use substitution to simplify the structure, then apply integration by parts to handle the product.

Key: Identify which method should be used first, depending on the main source of complexity in the function.

Example 1.1: Substitution First, Then Integration by Parts

Problem: Find ∫ln⁡xx2 dx\int \frac{\ln x}{x^2}\,dx

Solution:

Method 1: Direct Integration by Parts

Let u=ln⁡xu = \ln x, dv=1x2dxdv = \frac{1}{x^2}dx

Then du=1xdxdu = \frac{1}{x}dx, v=−1xv = -\frac{1}{x}

I=∫ln⁡xx2 dx=−ln⁡xx−∫−1x⋅1x dx=−ln⁡xx+∫1x2 dx=−ln⁡xx−1x+C=−ln⁡x+1x+C\begin{aligned} I &= \int \frac{\ln x}{x^2}\,dx = -\frac{\ln x}{x} - \int -\frac{1}{x} \cdot \frac{1}{x}\,dx \\ &= -\frac{\ln x}{x} + \int \frac{1}{x^2}\,dx = -\frac{\ln x}{x} - \frac{1}{x} + C \\ &= -\frac{\ln x + 1}{x} + C \end{aligned}

Method 2: First use substitution

Let t=ln⁡xt = \ln x, then x=etx = e^t, dx=etdtdx = e^t dt

I=∫te2t⋅et dt=∫te−t dtI = \int \frac{t}{e^{2t}} \cdot e^t\,dt = \int te^{-t}\,dt

Then use integration by parts; the result is the same.

Answer

I=−ln⁡x+1x+CI = -\frac{\ln x + 1}{x} + C
Example 1.2: Complex Combination Application

Problem: Find ∫ex dx\int e^{\sqrt{x}}\,dx

Solution:

Step 1: Substitute to eliminate the radical

Let t=xt = \sqrt{x}, then x=t2x = t^2, dx=2t dtdx = 2t\,dt

I=∫et⋅2t dt=2∫tet dtI = \int e^t \cdot 2t\,dt = 2\int te^t\,dt

Step 2: Integration by Parts

Let u=tu = t, dv=etdtdv = e^t dt

∫tet dt=tet−∫et dt=tet−et+C=et(t−1)+C\begin{aligned} \int te^t\,dt &= te^t - \int e^t\,dt = te^t - e^t + C = e^t(t-1) + C \end{aligned}

Step 3: Back Substitution

I=2et(t−1)+C=2ex(x−1)+CI = 2e^t(t-1) + C = 2e^{\sqrt{x}}(\sqrt{x}-1) + C

Answer

I=2ex(x−1)+CI = 2e^{\sqrt{x}}(\sqrt{x}-1) + C

2. Recursive Formula Method

Theorem 2.1: Establishing Recursive Formulas

For certain types of integrals, such as In=∫xnex dxI_n = \int x^n e^x\,dx or In=∫sin⁡nx dxI_n = \int \sin^n x\,dx, we can establish recursive relations between InI_n and In−1I_{n-1} or In−2I_{n-2}.

Example 2.1: Establishing a Recursive Formula

Problem: Find the recursive formula for In=∫xnex dxI_n = \int x^n e^x\,dx

Solution:

Step 1: Integration by Parts

Let u=xnu = x^n, dv=exdxdv = e^x dx

In=xnex−∫nxn−1ex dx=xnex−n∫xn−1ex dx=xnex−nIn−1\begin{aligned} I_n &= x^n e^x - \int nx^{n-1} e^x\,dx \\ &= x^n e^x - n\int x^{n-1} e^x\,dx \\ &= x^n e^x - nI_{n-1} \end{aligned}

Step 2: Recursive Formula

In=xnex−nIn−1I_n = x^n e^x - nI_{n-1}

Step 3: Application Example

Initial value: I0=∫ex dx=ex+CI_0 = \int e^x\,dx = e^x + C

I1=xex−1⋅I0=xex−ex+C=ex(x−1)+CI2=x2ex−2I1=x2ex−2ex(x−1)+C=ex(x2−2x+2)+C\begin{aligned} I_1 &= xe^x - 1 \cdot I_0 = xe^x - e^x + C = e^x(x-1) + C \\ I_2 &= x^2e^x - 2I_1 = x^2e^x - 2e^x(x-1) + C = e^x(x^2-2x+2) + C \end{aligned}

Recursive Formula

In=xnex−nIn−1,I0=ex+CI_n = x^n e^x - nI_{n-1}, \quad I_0 = e^x + C

3. Integration Cancellation Techniques

Definition 3.1: Integration Cancellation Method

When the original integral reappears after integrating twice, set the original integral as II and establish an algebraic equation in terms of II to solve.

This method is commonly used for integrals of type eaxsin⁡bxe^{ax}\sin bx or eaxcos⁡bxe^{ax}\cos bx.

Example 3.1: Classic Cancellation Problem

Problem: Find I=∫excos⁡x dxI = \int e^x \cos x\,dx

Solution:

Step 1: First integration by parts

Let u=exu = e^x, dv=cos⁡x dxdv = \cos x\,dx

I=exsin⁡x−∫exsin⁡x dxI = e^x \sin x - \int e^x \sin x\,dx

Step 2: Second integration by parts

For ∫exsin⁡x dx\int e^x \sin x\,dx, let u=exu = e^x, dv=sin⁡x dxdv = \sin x\,dx

∫exsin⁡x dx=ex(−cos⁡x)−∫ex(−cos⁡x) dx=−excos⁡x+∫excos⁡x dx=−excos⁡x+I\begin{aligned} \int e^x \sin x\,dx &= e^x(-\cos x) - \int e^x(-\cos x)\,dx \\ &= -e^x \cos x + \int e^x \cos x\,dx \\ &= -e^x \cos x + I \end{aligned}

Step 3: Substitute and solve

I=exsin⁡x−(−excos⁡x+I)I=exsin⁡x+excos⁡x−I2I=ex(sin⁡x+cos⁡x)I=ex(sin⁡x+cos⁡x)2+C\begin{aligned} I &= e^x \sin x - (-e^x \cos x + I) \\ I &= e^x \sin x + e^x \cos x - I \\ 2I &= e^x(\sin x + \cos x) \\ I &= \frac{e^x(\sin x + \cos x)}{2} + C \end{aligned}

Answer

I=ex(sin⁡x+cos⁡x)2+CI = \frac{e^x(\sin x + \cos x)}{2} + C
Example 3.2: More Complex Cancellation

Problem: Find I=∫e2xsin⁡3x dxI = \int e^{2x} \sin 3x\,dx

Solution:

Step 1: First integration by parts

I=13e2x(−cos⁡3x)+23∫e2xcos⁡3x dxI = \frac{1}{3}e^{2x}(-\cos 3x) + \frac{2}{3}\int e^{2x}\cos 3x\,dx

Step 2: Second integration by parts

∫e2xcos⁡3x dx=13e2xsin⁡3x−23∫e2xsin⁡3x dx\int e^{2x}\cos 3x\,dx = \frac{1}{3}e^{2x}\sin 3x - \frac{2}{3}\int e^{2x}\sin 3x\,dx

Step 3: Substitute and solve

I=−13e2xcos⁡3x+23(13e2xsin⁡3x−23I)I=−13e2xcos⁡3x+29e2xsin⁡3x−49I139I=e2x(29sin⁡3x−13cos⁡3x)I=e2x(2sin⁡3x−3cos⁡3x)13+C\begin{aligned} I &= -\frac{1}{3}e^{2x}\cos 3x + \frac{2}{3}\left(\frac{1}{3}e^{2x}\sin 3x - \frac{2}{3}I\right) \\ I &= -\frac{1}{3}e^{2x}\cos 3x + \frac{2}{9}e^{2x}\sin 3x - \frac{4}{9}I \\ \frac{13}{9}I &= e^{2x}\left(\frac{2}{9}\sin 3x - \frac{1}{3}\cos 3x\right) \\ I &= \frac{e^{2x}(2\sin 3x - 3\cos 3x)}{13} + C \end{aligned}

Answer

I=e2x(2sin⁡3x−3cos⁡3x)13+CI = \frac{e^{2x}(2\sin 3x - 3\cos 3x)}{13} + C

4. Denominator Reduction Method

Definition 4.1: Denominator Reduction Principle

For integrals of type ∫P(x)Q(x) dx\int \frac{P(x)}{Q(x)}\,dx, if direct integration is difficult, try using integration by parts to "reduce" the denominator (lower its degree).

Basic idea: Set I=uv−∫v duI = uv - \int v\,du, choose uu so that the denominator degree of vv is reduced.

Example 4.1: Reducing Denominator Degree

Problem: Find I=∫1(1+x2)2 dxI = \int \frac{1}{(1+x^2)^2}\,dx

Solution:

Step 1: Set up integration by parts

Write the integrand as 1(1+x2)2=11+x2⋅11+x2\frac{1}{(1+x^2)^2} = \frac{1}{1+x^2} \cdot \frac{1}{1+x^2}

Let u=11+x2u = \frac{1}{1+x^2}, dv=dxdv = dx

Then du=−2x(1+x2)2dxdu = -\frac{2x}{(1+x^2)^2}dx, v=xv = x

Step 2: Apply the formula

I=x1+x2−∫x⋅(−2x(1+x2)2)dx=x1+x2+2∫x2(1+x2)2 dx\begin{aligned} I &= \frac{x}{1+x^2} - \int x \cdot \left(-\frac{2x}{(1+x^2)^2}\right)dx \\ &= \frac{x}{1+x^2} + 2\int \frac{x^2}{(1+x^2)^2}\,dx \end{aligned}

Step 3: Simplify the second term

Notice that x2=(1+x2)−1x^2 = (1+x^2) - 1:

∫x2(1+x2)2 dx=∫(1+x2)−1(1+x2)2 dx=∫11+x2 dx−∫1(1+x2)2 dx=arctan⁡x−I\begin{aligned} \int \frac{x^2}{(1+x^2)^2}\,dx &= \int \frac{(1+x^2)-1}{(1+x^2)^2}\,dx \\ &= \int \frac{1}{1+x^2}\,dx - \int \frac{1}{(1+x^2)^2}\,dx \\ &= \arctan x - I \end{aligned}

Step 4: Solve for I

I=x1+x2+2(arctan⁡x−I)3I=x1+x2+2arctan⁡xI=x3(1+x2)+23arctan⁡x+C\begin{aligned} I &= \frac{x}{1+x^2} + 2(\arctan x - I) \\ 3I &= \frac{x}{1+x^2} + 2\arctan x \\ I &= \frac{x}{3(1+x^2)} + \frac{2}{3}\arctan x + C \end{aligned}

Answer

I=x3(1+x2)+23arctan⁡x+CI = \frac{x}{3(1+x^2)} + \frac{2}{3}\arctan x + C

5. Comprehensive Techniques Summary

Technique Selection Guide

1. Substitution First

When integrand contains radicals or complex composite functions

2. Integration by Parts

Product-type functions, especially products of polynomials and transcendental functions

3. Reduction Formulas

High-degree integrals with repeatedly appearing structures

4. Cancellation Techniques

Products of exponential and trigonometric functions

Problem-Solving Approach

Observe Structure

Identify the main source of complexity in the integrand

Attempt Simplification

Can identity transformations or substitutions eliminate complex terms?

Choose Method

Select the most suitable integration technique based on function type

Verify Answer

Check by differentiation to ensure correctness