Question

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

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

Hint:

Factorise the denominator as (4 + x)(1 - x), split into partial fractions and integrate each term.

Answer 1

The Correct Option is D

Step 1: Cover-up method
At the root $x=-4$ the other factor is $1-(-4)=5$, so $A=\frac{2(-4)}{5}=-\frac{8}{5}$. At $x=1$ the other factor is $5$, so $B=\frac{2}{5}$.

Step 2: Integrate each term
$\int \frac{dx}{1-x}=-\log |1-x|$, so the second term contributes $-\frac{2}{5}\log |1-x|$.

Step 3: Combine
The result is $-\frac{8}{5}\log |4+x|-\frac{2}{5}\log |1-x|+c$.

Final Answer:
The cover-up method confirms the same result. \[ \boxed{\text{(D) }-\dfrac{8}{5}\log |4+x|-\dfrac{2}{5}\log |1-x|+c} \]