Q1276: Find: $\int\frac{1}{x[(log~x)^{2}-3~log~x-4]}dx$...

Class CBSE Class 12 Mathematics Integrals Q #1276

KNOWLEDGE BASED

APPLY

3 Marks 2024 AISSCE(Board Exam) SA

Find: $\int\frac{1}{x[(log~x)^{2}-3~log~x-4]}dx$

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AI Tutor Explanation

Detailed Solution

Step 1: Substitution

Let $log~x = t$. Then, $\frac{1}{x}dx = dt$. The integral becomes: $\int \frac{1}{t^2 - 3t - 4} dt$

Step 2: Factorization

Factorize the denominator: $t^2 - 3t - 4 = (t-4)(t+1)$ So, the integral is now: $\int \frac{1}{(t-4)(t+1)} dt$

Step 3: Partial Fraction Decomposition

Use partial fraction decomposition: $\frac{1}{(t-4)(t+1)} = \frac{A}{t-4} + \frac{B}{t+1}$ $1 = A(t+1) + B(t-4)$

Step 4: Solving for A and B

To find $A$, let $t = 4$: $1 = A(4+1) + B(4-4) \implies 1 = 5A \implies A = \frac{1}{5}$ To find $B$, let $t = -1$: $1 = A(-1+1) + B(-1-4) \implies 1 = -5B \implies B = -\frac{1}{5}$ So, $\frac{1}{(t-4)(t+1)} = \frac{1}{5(t-4)} - \frac{1}{5(t+1)}$

Step 5: Integration

Now, integrate: $\int \frac{1}{5(t-4)} - \frac{1}{5(t+1)} dt = \frac{1}{5} \int \frac{1}{t-4} - \frac{1}{t+1} dt$ $= \frac{1}{5} [log|t-4| - log|t+1|] + C$ $= \frac{1}{5} log|\frac{t-4}{t+1}| + C$

Step 6: Substitute back

Substitute $t = log~x$: $\frac{1}{5} log|\frac{log~x - 4}{log~x + 1}| + C$

Final Answer: $\frac{1}{5} log|\frac{log~x - 4}{log~x + 1}| + C$

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Pedagogical Audit

Bloom's Analysis: This is an APPLY question because the student needs to apply the techniques of substitution, partial fraction decomposition, and integration to solve the problem.

Knowledge Dimension: PROCEDURAL

Justification: The question requires the student to apply a series of procedures like substitution, factorization, partial fraction decomposition, and integration.<\/span>

Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's knowledge of integration techniques, specifically substitution and partial fractions, which are standard topics in the syllabus.