Question:
If \( \int \frac{2x+5}{(x-1)(x+1)(x+4)(x+6)} \, dx = \frac{1}{10}\log\left(\frac{f(x)}{g(x)}\right)+c \) and \( \frac{f(-2)}{g(-2)}=6 \), then \( \frac{f(10)}{g(10)} = \)
If \( \int \frac{2x+5}{(x-1)(x+1)(x+4)(x+6)} \, dx = \frac{1}{10}\log\left(\frac{f(x)}{g(x)}\right)+c \) and \( \frac{f(-2)}{g(-2)}=6 \), then \( \frac{f(10)}{g(10)} = \)
Show Hint
Whenever the roots of the denominator can be grouped to produce a constant sum (\( -1+6 = 5 \) and \( 1+4 = 5 \)), it is a clear indicator that substituting \( t = x^2 + \text{sum} \cdot x \) will quickly simplify the expression.
Updated On: Jun 7, 2026
- \( \frac{72}{77} \)
- \( \frac{144}{75} \)
- \( \frac{55}{63} \)
- \( \frac{70}{59} \)
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The Correct Option is A
Solution and Explanation
Concept:
Let us group the factors in the denominator symmetrically to create a shared quadratic core term:
\[
(x-1)(x+6) = x^2 + 5x - 6
\]
\[
(x+1)(x+4) = x^2 + 5x + 4
\]
This grouping creates a matching term \( x^2 + 5x \) which simplifies the subsequent substitution method.
Step 1: Performing algebraic substitution.
Let \( t = x^2 + 5x \implies dt = (2x + 5)dx \). Substituting these into our integral expression: \[ I = \int \frac{1}{(t-6)(t+4)} dt \]
Step 2: Applying partial fractions.
The difference between the two linear factors in the denominator is exactly 10: \[ I = \frac{1}{10} \int \left( \frac{1}{t-6} - \frac{1}{t+4} \right) dt = \frac{1}{10} \log\left| \frac{t-6}{t+4} \right| + c \] Substituting back the value of \( t \): \[ I = \frac{1}{10} \log\left| \frac{x^2+5x-6}{x^2+5x+4} \right| + c \]
Step 3: Identifying functions \( f(x) \) and \( g(x) \).
Comparing with the given form, we find: \[ f(x) = x^2+5x-6 = (x-1)(x+6) \] \[ g(x) = x^2+5x+4 = (x+1)(x+4) \]
Step 4: Evaluating the required ratio at \( x = 10 \).
Substitute \( x = 10 \): \[ f(10) = (10-1)(10+6) = 9 \times 16 = 144 \] \[ g(10) = (10+1)(10+4) = 11 \times 14 = 154 \] Calculating their simplified fractional ratio: \[ \frac{f(10)}{g(10)} = \frac{144}{154} = \frac{72}{77} \] This matches option (A) perfectly.
Step 1: Performing algebraic substitution.
Let \( t = x^2 + 5x \implies dt = (2x + 5)dx \). Substituting these into our integral expression: \[ I = \int \frac{1}{(t-6)(t+4)} dt \]
Step 2: Applying partial fractions.
The difference between the two linear factors in the denominator is exactly 10: \[ I = \frac{1}{10} \int \left( \frac{1}{t-6} - \frac{1}{t+4} \right) dt = \frac{1}{10} \log\left| \frac{t-6}{t+4} \right| + c \] Substituting back the value of \( t \): \[ I = \frac{1}{10} \log\left| \frac{x^2+5x-6}{x^2+5x+4} \right| + c \]
Step 3: Identifying functions \( f(x) \) and \( g(x) \).
Comparing with the given form, we find: \[ f(x) = x^2+5x-6 = (x-1)(x+6) \] \[ g(x) = x^2+5x+4 = (x+1)(x+4) \]
Step 4: Evaluating the required ratio at \( x = 10 \).
Substitute \( x = 10 \): \[ f(10) = (10-1)(10+6) = 9 \times 16 = 144 \] \[ g(10) = (10+1)(10+4) = 11 \times 14 = 154 \] Calculating their simplified fractional ratio: \[ \frac{f(10)}{g(10)} = \frac{144}{154} = \frac{72}{77} \] This matches option (A) perfectly.
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