Question:
\(\int \frac{x+2}{x^2-7x+12}dx\) is equal to
\(\int \frac{x+2}{x^2-7x+12}dx\) is equal to
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Factor the denominator as $(x-3)(x-4)$ and split into partial fractions.
Updated On: Oct 1, 2026
- \(-5log|x-3|+6log|x-4|+c\)
- \(-5log|x-3|-6log|x-4|+c\)
- \(5log|x-3|-6log|x-4|+c\)
- \(5log|x-3|+6log|x-4|+c\)
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The Correct Option is A
Solution and Explanation
Step 1: Factor
\(x^2-7x+12=(x-3)(x-4)\).
Step 2: Partial fractions
Write \(\frac{x+2}{(x-3)(x-4)}=\frac{A}{x-3}+\frac{B}{x-4}\).
Put \(x=3\): \(A=\frac{5}{-1}=-5\). Put \(x=4\): \(B=\frac{6}{1}=6\).
Step 3: Integrate
\[ \int\left(\frac{-5}{x-3}+\frac{6}{x-4}\right)dx=-5\log|x-3|+6\log|x-4|+c \]
Final Answer:
The integral is \(-5\log|x-3|+6\log|x-4|+c\), option (A). \[ \boxed{\text{(A)}} \]
\(x^2-7x+12=(x-3)(x-4)\).
Step 2: Partial fractions
Write \(\frac{x+2}{(x-3)(x-4)}=\frac{A}{x-3}+\frac{B}{x-4}\).
Put \(x=3\): \(A=\frac{5}{-1}=-5\). Put \(x=4\): \(B=\frac{6}{1}=6\).
Step 3: Integrate
\[ \int\left(\frac{-5}{x-3}+\frac{6}{x-4}\right)dx=-5\log|x-3|+6\log|x-4|+c \]
Final Answer:
The integral is \(-5\log|x-3|+6\log|x-4|+c\), option (A). \[ \boxed{\text{(A)}} \]
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