Math Problem Statement

5 ln(x2 − 1) dx

Solution

To solve the integral:

5ln(x21)dx\int 5 \ln(x^2 - 1) \, dx

we can use integration by parts. The formula for integration by parts is:

udv=uvvdu\int u \, dv = u v - \int v \, du

  1. Let u=ln(x21)u = \ln(x^2 - 1), so du=2xx21dxdu = \frac{2x}{x^2 - 1} \, dx.
  2. Let dv=5dxdv = 5 \, dx, so v=5xv = 5x.

Now, applying the formula:

5ln(x21)dx=5xln(x21)5x2xx21dx\int 5 \ln(x^2 - 1) \, dx = 5x \ln(x^2 - 1) - \int 5x \cdot \frac{2x}{x^2 - 1} \, dx

Simplifying the second integral:

10x2x21dx\int \frac{10x^2}{x^2 - 1} \, dx

Next, we split the fraction:

(10+10x21)dx=10x+101x21dx\int \left( 10 + \frac{10}{x^2 - 1} \right) \, dx = 10x + 10 \int \frac{1}{x^2 - 1} \, dx

The integral of 1x21\frac{1}{x^2 - 1} can be expressed as a difference of logarithms:

1x21dx=12lnx1x+1\int \frac{1}{x^2 - 1} \, dx = \frac{1}{2} \ln \left| \frac{x - 1}{x + 1} \right|

Now we combine everything:

5ln(x21)dx=5xln(x21)10x5lnx1x+1+C\int 5 \ln(x^2 - 1) \, dx = 5x \ln(x^2 - 1) - 10x - 5 \ln \left| \frac{x - 1}{x + 1} \right| + C

where CC is the constant of integration.


Would you like more details, or do you have any questions?

Here are 5 related questions to explore:

  1. How is integration by parts applied to logarithmic functions in other cases?
  2. Can you use substitution to solve integrals like this?
  3. How would the integral change if the argument of the logarithm was different, say ln(x2+1)\ln(x^2 + 1)?
  4. What are the applications of logarithmic integrals in physics or engineering?
  5. How can partial fraction decomposition help in solving rational integrals?

Tip: When facing complex logarithmic integrals, remember that splitting the logarithmic expression into simpler terms can often simplify the integration process.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Integration
Logarithmic Functions
Integration by Parts
Partial Fractions

Formulas

Integration by Parts: ∫u dv = uv − ∫v du
ln(x^2 − 1) = ln((x − 1)(x + 1))
∫(1/(x^2 − 1)) dx = (1/2) ln| (x − 1)/(x + 1) |

Theorems

Integration by Parts Theorem
Logarithmic Properties
Partial Fraction Decomposition

Suitable Grade Level

Undergraduate - Calculus II