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- Calculus
- Limits
- Continuity
- Definition of the Derivative - First Principles
- Basic Differentiation Rules
- More Differentiation Rules
- Implicit Differentiation
- Higher Order Derivatives
- Curve Sketching
- First Derivative Test
- Second Derivative Test
- Derivatives of Trigonometric Functions
- Derivatives of Inverse Trigonometric Functions
- Derivatives of Logarithmic Functions
- Derivatives of Exponential Functions
- Antiderivative
- Indefinite Integral
- Applications of the Indefinite Integral
- The Definite Integral
- Area Under a Curve - Riemann Sums
- Area Under a Curve - Integration
- Area between Two Curves
- Volumes of Revolution
- Logarithmic Integrals
- Exponential Integrals
- Trigonometric Integrals
- Trigonometric Integrals of Other Forms
- More Integration Methods

The power rule for integration is valid for all values except when the exponent is equal to \(-1\). This is because \[\frac{d(\ln u)}{dx} = \frac{1}{u}\frac{du}{dx}\]

Thus, reversing the process where the denominator's exponent is \(-1\) would lead to an integral of the logarithmic form.

\[\int\frac{1}{u}du=\ln |u|+C\] |

Thus, the integration of \[\int \frac{x \,dx}{3+x^2}\] will be done using the logarithmic form, whereas \[\int \frac{x\, dx}{(3+x^2)^2}\] will be done using the power rule for integration.

Example 1: Integrate the function \[\int_{1}^{2}\frac{1}{8-3x}dx\]

Solution: We can recognize this is an integral of logarithmic form because the denominator is to the power of -1 (e.g., it can be written as \((8-3x)^{-1}\).

Let \(u=8-3x\), \(du=-3dx\). We can substitute these values and change the variable to u

\[\frac{1}{-3}\int_{1}^{2}\frac{1}{u}du=\frac{1}{-3}\left[\ln |u|\right]_{1}^{2}\]

Solving for the definite integral we get,

\[\frac{\ln 2}{-3}-\frac{\ln 1}{-3} = \frac{\ln 2}{-3}\qquad \qquad (\ln 1=0)\]

Example 2: Integrate the function \[\int\frac{dx}{x(1+2\ln x)}\]

Solution: Let \(u=1+2\ln x\)

\begin{align} du &= \frac{2}{x}dx \\ \frac{dx}{x} &= \frac{du}{2} \end{align}

The integral after subitution is

\begin{align} &\int \frac{du}{2u} \\ &=\frac{\ln |u|}{2} +C \\ &=\frac{\ln |1+2\ln x|}{2} +C \end{align}

Example 3: Integrate \[\int \sec x\,dx\]

Solution: There is a trick here and that is to multiply \(\sec x\) by \[\frac{\sec x + \tan x}{\sec x + \tan x}\]

The resulting integral is

\[\int \frac{\sec ^2x + \sec x\tan x}{\sec x + \tan x}dx\]

Now let \(u=\sec x + \tan x\)

\[ du=\sec x\tan x + \sec ^2x \,dx\]

The integral becomes,

\[\int \frac{du}{u} = \ln |u| + C = \ln |\sec x + \tan x| + C \]

Designed by Matthew Cheung. This work is licensed under a Creative Commons Attribution 4.0 International License.

- Last Updated: Aug 12, 2022 11:32 AM
- URL: https://libraryguides.centennialcollege.ca/c.php?g=717032
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