Class CBSE Class 12 Mathematics Integrals Q #621
KNOWLEDGE BASED
APPLY
1 Marks 2024 AISSCE(Board Exam) MCQ SINGLE
\(\int\frac{1}{x(\log~x)^{2}}dx\) is equal to:
(A) \(2~log(log~x)+c\)
(B) \(-\frac{1}{log~x}+c\)
(C) \(\frac{(log~x)^{3}}{3}+c\)
(D) \(\frac{3}{(log~x)^{3}}+c\)
Prev Next

AI Tutor Explanation

Powered by Gemini

Step-by-Step Solution

Let \(I = \int\frac{1}{x(\log~x)^{2}}dx\)

Let \(u = \log~x\), then \(\frac{du}{dx} = \frac{1}{x}\), so \(du = \frac{1}{x}dx\)

Substituting into the integral, we get: \(I = \int\frac{1}{u^{2}}du = \int u^{-2}du\)

Using the power rule for integration, \(\int u^{n}du = \frac{u^{n+1}}{n+1} + c\), we have: \(I = \frac{u^{-2+1}}{-2+1} + c = \frac{u^{-1}}{-1} + c = -\frac{1}{u} + c\)

Substituting back \(u = \log~x\), we get: \(I = -\frac{1}{\log~x} + c\)

Correct Answer: -\frac{1}{log~x}+c

AI Suggestion: Option B

AI generated content. Review strictly for academic accuracy.

Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student needs to apply the knowledge of integration techniques, specifically substitution, to solve the given integral.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to execute a specific procedure (integration using substitution) to arrive at the solution. It involves knowing the steps and applying them correctly.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's ability to apply integration techniques learned from the textbook.

More from this Chapter

SA
(a) Evaluate: $\int_0^{2\pi} \frac{1}{1 + e^{\sin x}} dx $ OR (b) Find: $\int \frac{x⁴} { ((x-1)(x²+1))}dx.$
SA
Evaluate $\int_{1}^{e}\frac{1}{\sqrt{4x^{2}-(x\log x)^{2}}}dx$
SA
Evaluate : $\int_{0}^{\frac{\pi}{2}}e^{x }\sin x~dx$. OR Find: $\int\frac{1}{\cos(x-a)\cos(x-b)}dx$
MCQ_SINGLE
\(\int\frac{\cos 2x-\cos 2\alpha}{\cos x-\cos \alpha}dx\) is equal to:
SA
Evaluate $\int_{0}^{\frac{\pi}{2}}[\log(\sin~x)-\log(2\cos~x)]dx.$
View All Questions