Question

If $\int \frac{dx}{32 - 2x^2} = A \log|4 - x| + B \log|4 + x| + c$, then the values of $A$ and $B$ are respectively

• A. $-\frac{1}{8}, \frac{1}{8}$
• B. $\frac{1}{8}, -\frac{1}{8}$
• C. $-\frac{1}{16}, \frac{1}{16}$
• D. $\frac{1}{16}, \frac{1}{16}$

Hint:

You can verify partial fraction constants using differentiation! Take the derivative of the target format: $\frac{d}{dx}[A\log(4-x) + B\log(4+x)] = \frac{-A}{4-x} + \frac{B}{4+x} = \frac{A}{x-4} + \frac{B}{x+4}$. Setting this equal to the initial integrand $\frac{1}{2(4-x)(4+x)}$ lets you determine the matching values for $A$ and $B$ instantly.

Answer 1

The Correct Option is C

Step 1: Set up the integral.
We need $\displaystyle\int \dfrac{dx}{32-2x^2}$ matched to $A\log|4-x| + B\log|4+x| + c$.
Step 2: Factor the denominator.
$32 - 2x^2 = 2(16 - x^2) = 2(4-x)(4+x)$.
Step 3: Split into partial fractions.
Write $\dfrac{1}{2(4-x)(4+x)} = \dfrac{P}{4-x} + \dfrac{Q}{4+x}$, so $\dfrac{1}{2} = P(4+x) + Q(4-x)$.
Step 4: Solve for $P$ and $Q$.
Put $x=4$: $\dfrac{1}{2} = 8P$, so $P = \dfrac{1}{16}$. Put $x=-4$: $\dfrac{1}{2} = 8Q$, so $Q = \dfrac{1}{16}$.
Step 5: Integrate each piece.
$\displaystyle\int \dfrac{1/16}{4-x}dx = -\dfrac{1}{16}\log|4-x|$ (the minus comes from the $-x$), and $\displaystyle\int \dfrac{1/16}{4+x}dx = \dfrac{1}{16}\log|4+x|$.
Step 6: Compare coefficients.
So $A = -\dfrac{1}{16}$ and $B = \dfrac{1}{16}$, the same result the standard formula gives.
\[ \boxed{A = -\dfrac{1}{16},\ B = \dfrac{1}{16}} \]