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Q14:
Integrate the functions in Exercises 1 to 37: 14. $\frac{1}{x(\log x)^m}$, $x > 0, m \neq 1$

Solution :

Step 1: Formal Representation of the Integral

We are required to evaluate the indefinite integral denoted by $I$: $$I = \int \frac{1}{x(\log x)^m} \, dx$$ [Given the domain constraints $x > 0$ and $m \neq 1$ to ensure the function is defined and the integral does not result in a logarithmic singularity.]

Step 2: Substitution Methodology

To simplify the integrand, we perform a change of variable. Let:

  • $u = \log x$
  • By applying the differential operator, we obtain: $\frac{du}{dx} = \frac{1}{x}$
  • Rearranging for the differential element: $du = \frac{1}{x} \, dx$

[By the Method of Substitution, where we replace a composite function part with a single variable to simplify the integrand.]

Step 3: Transforming the Integral

Substituting $u = \log x$ and $du = \frac{1}{x} \, dx$ into the original integral expression:

$$I = \int \frac{1}{u^m} \, du$$

Rewriting the integrand using the laws of exponents: $$I = \int u^{-m} \, du$$

Step 4: Integration Using the Power Rule

Applying the standard integration power rule, $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ (where $n \neq -1$):

$$I = \frac{u^{-m+1}}{-m+1} + C$$

Simplifying the exponent and the denominator: $$I = \frac{u^{1-m}}{1-m} + C$$

Step 5: Re-substitution and Final Expression

Substitute $u = \log x$ back into the expression to return to the original variable $x$:

$$I = \frac{(\log x)^{1-m}}{1-m} + C$$ Integral path: f(x) = 1/(x(log x)^m) Substitution: u = log(x) Integration: ∫ u^-m du = u^(1-m)/(1-m)

Final Solution: \frac{(\log x)^{1-m}}{1-m} + C


More Questions from Class 12 Mathematics Integrals EXERCISE 7.2


CBSE Solutions for Class 12 Mathematics Integrals


Chapters in CBSE - Class 12 Mathematics


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