Math Problem Statement

Integrate 1/(logx)^2

Solution

To solve the integral of 1(logx)2\frac{1}{(\log{x})^2} with respect to xx, we can apply a substitution method.

Step-by-Step Solution:

  1. Substitution: Let u=logxu = \log{x}.
    Therefore, du=1xdxdu = \frac{1}{x} dx, which gives us dx=xdudx = x \, du.

    Since u=logxu = \log{x}, we also have x=eux = e^u. Thus: dx=eududx = e^u \, du

  2. Rewrite the Integral: Substituting these into the integral, we get: 1(logx)2dx=1u2eudu\int \frac{1}{(\log{x})^2} \, dx = \int \frac{1}{u^2} \cdot e^u \, du

  3. Integrating by Parts: Now, we integrate euu2\frac{e^u}{u^2} using integration by parts.

    Let v=1u2v = \frac{1}{u^2} and dw=eududw = e^u du.

    So, we apply integration by parts: vdw=vwwdv\int v \, dw = v \cdot w - \int w \, dv This method simplifies the integral, yielding: euu2du=euu+Ei(u)\int \frac{e^u}{u^2} \, du = -\frac{e^u}{u} + \text{Ei}(u) where Ei(u)\text{Ei}(u) is the exponential integral function.

  4. Substitute Back: Substituting back u=logxu = \log{x}, we get: 1(logx)2dx=1logx+Ei(logx)+C\int \frac{1}{(\log{x})^2} \, dx = -\frac{1}{\log{x}} + \text{Ei}(\log{x}) + C where CC is the constant of integration.

Final Answer:

1logx+Ei(logx)+C\boxed{-\frac{1}{\log{x}} + \text{Ei}(\log{x}) + C}

Let me know if you'd like any clarification or additional details.


Related Questions:

  1. How can we use the substitution method for integrals with different logarithmic expressions?
  2. What is the definition and properties of the Exponential Integral function Ei(x)\text{Ei}(x)?
  3. Can we solve 1(logx)ndx\int \frac{1}{(\log{x})^n} \, dx for other values of nn?
  4. How do we approach integrals involving products of exponentials and powers of logarithms?
  5. What are the applications of logarithmic integrals in calculus?

Tip:

When solving integrals involving logarithms, substitution is often useful to simplify the expression, especially when combined with integration by parts.

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Math Problem Analysis

Mathematical Concepts

Integration
Logarithms
Substitution Method
Integration by Parts

Formulas

∫ 1/(logx)^2 dx
Substitution: u = log(x), du = 1/x dx
Integration by parts formula: ∫ v dw = v w - ∫ w dv

Theorems

Exponential Integral Function (Ei)
Logarithmic Properties

Suitable Grade Level

Undergraduate Calculus