CBSE Class 12 Mathematics Integrals Q #1777
COMPETENCY BASED
APPLY
3 Marks 2026 AISSCE(Board Exam) SA
Find: $\int\frac{x+2}{\sqrt{9x-x^{2}}}dx$
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Detailed Solution

Step 1: Express the numerator in terms of the derivative of the quadratic expression

Let the numerator $x+2 = A \frac{d}{dx}(9x-x^2) + B$. The derivative of $9x-x^2$ is $9-2x$. So, $x+2 = A(9-2x) + B$.

Step 2: Solve for constants A and B

Comparing coefficients of $x$: $-2A = 1 \implies A = -1/2$. Comparing constants: $9A + B = 2 \implies 9(-1/2) + B = 2 \implies B = 2 + 4.5 = 13/2$. Thus, $x+2 = -\frac{1}{2}(9-2x) + \frac{13}{2}$.

Step 3: Split the integral

The integral becomes: $$I = -\frac{1}{2} \int \frac{9-2x}{\sqrt{9x-x^2}} dx + \frac{13}{2} \int \frac{1}{\sqrt{9x-x^2}} dx$$

Step 4: Evaluate the integrals

For the first part, use substitution $u = 9x-x^2$, $du = (9-2x)dx$: $$I_1 = -\frac{1}{2} \int u^{-1/2} du = -\frac{1}{2} (2\sqrt{u}) = -\sqrt{9x-x^2}$$ For the second part, complete the square: $9x-x^2 = -(x^2-9x) = -((x-9/2)^2 - 81/4) = (9/2)^2 - (x-9/2)^2$. $$I_2 = \frac{13}{2} \int \frac{dx}{\sqrt{(9/2)^2 - (x-9/2)^2}} = \frac{13}{2} \sin^{-1}\left(\frac{x-9/2}{9/2}\right) = \frac{13}{2} \sin^{-1}\left(\frac{2x-9}{9}\right)$$

Final Answer: -\sqrt{9x-x^2} + \frac{13}{2} \sin^{-1}\left(\frac{2x-9}{9}\right) + C

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student must apply the standard algebraic technique of splitting the numerator to match the derivative of the denominator, followed by standard integration formulas.
Knowledge Dimension: PROCEDURAL
Justification: The question requires a specific sequence of algorithmic steps (completing the square and substitution) to reach the solution.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. It tests the student's ability to manipulate expressions to fit standard integral forms, a core competency in the Calculus unit.

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