1 5-2x 2 Calculate Integration by Subsitution

This guide explains how to calculate the integral of 1/(5-2x) using substitution. We'll cover the method, provide a step-by-step solution, and include a built-in calculator to perform the calculation quickly.

How to Calculate the Integral of 1/(5-2x)

The integral of 1/(5-2x) can be solved using substitution. This method involves transforming the integrand into a simpler form that can be integrated more easily. Here's an overview of the process:

Formula: ∫(1/(5-2x)) dx = -1/2 * ln|5-2x| + C

The key steps are:

Identify the substitution variable (u)
Express the differential (du)
Rewrite the integral in terms of u
Integrate with respect to u
Substitute back to the original variable

This method is particularly useful for rational functions where the numerator is the derivative of the denominator.

Step-by-Step Guide

Step 1: Identify the substitution

Let u = 5 - 2x. This choice is based on the denominator of the integrand.

Step 2: Differentiate to find du

Differentiate both sides with respect to x:

du/dx = -2

du = -2 dx

Step 3: Rewrite the integral

Express dx in terms of du:

dx = -1/2 du

Substitute into the original integral:

∫(1/(5-2x)) dx = ∫(1/u) (-1/2 du)

Step 4: Integrate with respect to u

The integral of 1/u is ln|u|:

∫(1/u) (-1/2 du) = -1/2 ∫(1/u) du = -1/2 ln|u| + C

Step 5: Substitute back to x

Replace u with 5 - 2x:

-1/2 ln|5-2x| + C

Example Calculation

Let's calculate the definite integral from x=0 to x=1:

∫[0,1] (1/(5-2x)) dx = [-1/2 ln|5-2x|][0,1]

= -1/2 [ln|5-2(1)| - ln|5-2(0)|]

= -1/2 [ln|3| - ln|5|]

= -1/2 [ln(3/5)]

= -1/2 ln(0.6)

The result is approximately -0.2027.

Note: The natural logarithm of a negative number is not defined in real numbers. The absolute value ensures the argument is positive.

Common Mistakes to Avoid

Forgetting to include the absolute value in the logarithm
Incorrectly differentiating the substitution variable
Miscounting the sign when solving for dx in terms of du
Omitting the constant of integration (C) in indefinite integrals
Misapplying the limits of integration in definite integrals

FAQ

What is the integral of 1/(5-2x)?: The integral of 1/(5-2x) is -1/2 ln|5-2x| + C.
Can I use substitution for any rational function?: Substitution works well when the numerator is the derivative of the denominator, as in this case.
What happens if the denominator becomes zero?: The integrand becomes undefined, and the integral has a vertical asymptote at that point.
Is there a simpler method for this integral?: Substitution is the most straightforward method for this particular integral.
How do I handle definite integrals with this function?: Apply the antiderivative at the upper and lower limits, then subtract the results.