Question:
\(\int \frac{2x}{4-3x-x^{2}}\ dx=\ldots\)
\(\int \frac{2x}{4-3x-x^{2}}\ dx=\ldots\)
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Factorise the denominator as (4 + x)(1 - x), split into partial fractions and integrate each term.
Updated On: Nov 12, 2026
- \(-\frac{8}{5}\log |4-x|-\frac{2}{5}\log |1+x|+c\)
- \(-\frac{2}{5}\log |4+x|-\frac{8}{5}\log |1-x|+c\)
- \(-\frac{2}{5}\log |4-x|-\frac{8}{5}\log |1+x|+c\)
- \(-\frac{8}{5}\log |4+x|-\frac{2}{5}\log |1-x|+c\)
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The Correct Option is D
Solution and Explanation
Step 1: Factorise
\(4-3x-x^{2}=-(x+4)(x-1)=(4+x)(1-x)\).
Step 2: Split into partial fractions
\(\frac{2x}{(4+x)(1-x)}=\frac{A}{4+x}+\frac{B}{1-x}\), so \(2x=A(1-x)+B(4+x)\).
Step 3: Find A and B
Put \(x=1\): \(2=5B\Rightarrow B=\frac{2}{5}\).
Put \(x=-4\): \(-8=5A\Rightarrow A=-\frac{8}{5}\).
Step 4: Integrate
\(\int \left[-\frac{8}{5}\frac{1}{4+x}+\frac{2}{5}\frac{1}{1-x}\right]dx=-\frac{8}{5}\log |4+x|-\frac{2}{5}\log |1-x|+c\).
Final Answer:
The integral is \(-\frac{8}{5}\log |4+x|-\frac{2}{5}\log |1-x|+c\). \[ \boxed{\text{(D) }-\dfrac{8}{5}\log |4+x|-\dfrac{2}{5}\log |1-x|+c} \]
\(4-3x-x^{2}=-(x+4)(x-1)=(4+x)(1-x)\).
Step 2: Split into partial fractions
\(\frac{2x}{(4+x)(1-x)}=\frac{A}{4+x}+\frac{B}{1-x}\), so \(2x=A(1-x)+B(4+x)\).
Step 3: Find A and B
Put \(x=1\): \(2=5B\Rightarrow B=\frac{2}{5}\).
Put \(x=-4\): \(-8=5A\Rightarrow A=-\frac{8}{5}\).
Step 4: Integrate
\(\int \left[-\frac{8}{5}\frac{1}{4+x}+\frac{2}{5}\frac{1}{1-x}\right]dx=-\frac{8}{5}\log |4+x|-\frac{2}{5}\log |1-x|+c\).
Final Answer:
The integral is \(-\frac{8}{5}\log |4+x|-\frac{2}{5}\log |1-x|+c\). \[ \boxed{\text{(D) }-\dfrac{8}{5}\log |4+x|-\dfrac{2}{5}\log |1-x|+c} \]
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