Understanding and Solving the Integral of 1/x^2 * ln x

The integral of the form ( int frac{1}{x^2 ln x} , dx ) can be approached in multiple ways depending on how the logarithm is structured within the integral. This article will explore two distinct methodologies, one where the logarithm is treated as part of the denominator, and another where it is part of the numerator. Let's dive into these methods in detail.

Method 1: ln(x) in the Denominator

To begin, let us rewrite the integral for the case where ln(x) is in the denominator:

$$I int frac{dx}{x^2 ln x}$$

We will use the substitution (x e^t) to simplify the integral. This substitution yields:

$$dx e^t , dt$$

Substituting these values into the integral, we get:

$$I int frac{e^t , dt}{e^{2t} t} int frac{e^{-t} , dt}{t}$$

The integral of this form is related to the exponential integral function, denoted as (E_1(t)). Using this function, the integral can be expressed as:

$$I -E_1(t) C$$

Since (t ln x), we replace (t) with (ln x) to obtain the final expression:

$$I -E_1(ln x) C$$

Here, (E_1(ln x)) represents the exponential integral evaluated at (ln x).

Method 2: ln(x) in the Numerator

Now, let's consider the case where ln(x) is in the numerator:

$$P int frac{ln x}{x^2} , dx$$

To solve this integral, we can use integration by parts. Let us take (u ln x) and (dv frac{1}{x^2} , dx). This yields:

Choose:

$$u ln x, quad du frac{1}{x} , dx$$

$$dv frac{1}{x^2} , dx, quad v -frac{1}{x}$$

Using the integration by parts formula ( int u , dv uv - int v , du ), we have:

$$P int frac{ln x}{x^2} , dx -frac{ln x}{x} int frac{1}{x^2} , dx$$

The remaining integral is straightforward:

$$int frac{1}{x^2} , dx -frac{1}{x} C$$

Substituting this back into the expression for (P), we get:

$$P -frac{ln x 1}{x} C$$

Thus, the value of the integral is:

$$P -frac{ln x 1}{x} C$$

Conclusion

Understanding and applying the correct method to solve integrals involving logarithms can significantly streamline the problem-solving process. Both methods we discussed are valid, and the choice between them depends on the specific structure of the integral. Whether you're dealing with ln(x) in the denominator or the numerator, the key is to recognize the form and apply the appropriate substitution or integration technique.