Question

If \[ \int \frac{dx}{x^{2022}(1+x^{2022})^{\frac{1}{2022}}} = -\frac{(1+x^m)^{\frac{n}{m}}}{nx^n}+C, \] then \(m-n=\)

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(0\)

Hint:

For integrals of the form \[ \int \frac{dx}{x^{n+1}(1+x^m)^{1-\frac{n}{m}}}, \] use the result \[ \int \frac{dx}{x^{n+1}(1+x^m)^{1-\frac{n}{m}}} = -\frac{(1+x^m)^{\frac{n}{m}}}{nx^n}+C \] and compare powers carefully.

Answer 1

The Correct Option is A

Step 1: Plan - differentiate the RHS to match the integrand.
Given $ -\frac{(1+x^m)^{n/m}}{n\cdot x^n}+C $, differentiate and match to $ \frac{1}{x^{2022}(1+x^{2022})^{1/2022}} $.
Step 2: Differentiate $ F=-\frac{1}{n}(1+x^m)^{n/m}\cdot x^{-n} $ by the product rule.
\[ F'=-\frac{1}{n}\!\left[\frac{n}{m}mx^{m-1}(1+x^m)^{n/m-1}\cdot x^{-n}+(1+x^m)^{n/m}(-n)x^{-n-1}\right] \] \[ = x^{-n-1}(1+x^m)^{n/m-1}\!\left[-x^m+(1+x^m)\right] = x^{-n-1}(1+x^m)^{n/m-1} \]
Step 3: Match the exponent of x.
$ -n-1=-2022 \implies n=2021 $.
Step 4: Match the exponent of $ (1+x^m) $.
$ \frac{n}{m}-1=-\frac{1}{2022} \implies \frac{2021}{m}=\frac{2021}{2022} \implies m=2022 $.
Step 5: Verify the match.
$ F'=x^{-2022}(1+x^{2022})^{-1/2022} $. This equals the integrand exactly.
Step 6: Compute $ m-n $.
\[ m-n=2022-2021=1 \] \[ \boxed{1} \]