Question

\(\int \frac{2x}{4-3x-x^2}dx = \cdots\)

• A. \(\frac{8}{5}log|4+x|-\frac{2}{5}log|1-x|+c\)
• B. \(2log|4+x|+\frac{2}{5}log|1-x|+c\)
• C. \(-log|4+x|+\frac{2}{5}log|1-x|+c\)
• D. \(-\frac{8}{5}log|4+x|-\frac{2}{5}log|1-x|+c\)

Hint:

Factor the denominator as (1 - x)(4 + x) and use partial fractions.

Answer 1

The Correct Option is D

Step 1: Factor the denominator:
\(4-3x-x^2=-(x^2+3x-4)=-(x+4)(x-1)=(1-x)(4+x)\).

Step 2: Partial fractions:
\(\dfrac{2x}{(1-x)(4+x)}=\dfrac{A}{1-x}+\dfrac{B}{4+x}\), so \(2x=A(4+x)+B(1-x)\). Put \(x=1\): \(2=5A\), so \(A=\dfrac25\). Put \(x=-4\): \(-8=5B\), so \(B=-\dfrac85\).

Step 3: Integrate:
\(\int\dfrac{A}{1-x}dx=-A\ln|1-x|\) and \(\int\dfrac{B}{4+x}dx=B\ln|4+x|\).
\[ I=-\dfrac25\ln|1-x|-\dfrac85\ln|4+x|+c \]

Step 4: Why the other options are wrong.
Options A, B and C have wrong coefficients or a wrong sign on the \(\ln|1-x|\) term, which comes from forgetting the minus sign when integrating \(\dfrac{1}{1-x}\).

Final Answer:
The integral is -(8/5) log|4 + x| - (2/5) log|1 - x| + c. \[ \boxed{\text{(D) }-\dfrac85\log|4+x|-\dfrac25\log|1-x|+c} \]