Class CBSE Class 12 Mathematics Integrals Q #1338
KNOWLEDGE BASED
UNDERSTAND
2 Marks 2024 AISSCE(Board Exam) VSA
Find: $\int\frac{1}{5+4x-x^{2}}dx$
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Detailed Solution

Step 1: Complete the square in the denominator

We have $5 + 4x - x^2$. To complete the square, we rewrite it as: $5 - (x^2 - 4x)$. Now, we add and subtract $(4/2)^2 = 4$ inside the parenthesis: $5 - (x^2 - 4x + 4 - 4) = 5 - ((x - 2)^2 - 4) = 5 - (x - 2)^2 + 4 = 9 - (x - 2)^2$.

Step 2: Rewrite the integral

Now we can rewrite the integral as: $\int \frac{1}{9 - (x - 2)^2} dx = \int \frac{1}{3^2 - (x - 2)^2} dx$.

Step 3: Apply the standard integral formula

We use the formula $\int \frac{1}{a^2 - x^2} dx = \frac{1}{2a} \ln \left| \frac{a + x}{a - x} \right| + C$. In our case, $a = 3$ and $x$ is replaced by $(x - 2)$. So, $\int \frac{1}{3^2 - (x - 2)^2} dx = \frac{1}{2(3)} \ln \left| \frac{3 + (x - 2)}{3 - (x - 2)} \right| + C = \frac{1}{6} \ln \left| \frac{1 + x}{5 - x} \right| + C$.

Step 4: Final Answer

Therefore, the integral is: $\int \frac{1}{5 + 4x - x^2} dx = \frac{1}{6} \ln \left| \frac{x + 1}{5 - x} \right| + C$.

Final Answer: $\frac{1}{6} \ln \left| \frac{x+1}{5-x} \right| + C$

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Pedagogical Audit
Bloom's Analysis: This is an UNDERSTAND question because the student needs to understand the process of completing the square and applying the standard integral formula.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to apply a specific procedure (completing the square and using a standard integral formula) to solve the problem.
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 and standard formulas as covered in the textbook.

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