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∫₀¹ 2/(x³+6x²+11x+6) dx Definite | JEE

JEE Maths question with a full step-by-step solution.

Question
∫012x3+6x2+11x+6 dx\displaystyle\int_{0}^{1}\dfrac{2}{x^{3}+6x^{2}+11x+6}\,dx equals
A5ln⁡2−3ln⁡35\ln 2-3\ln 3correct
B3ln⁡3−5ln⁡23\ln 3-5\ln 2
C5ln⁡2+3ln⁡35\ln 2+3\ln 3
D5ln⁡5−2ln⁡25\ln 5-2\ln 2
Solution
Step 1: Factor the cubic. The roots x=−1,−2,−3x=-1,-2,-3 each annihilate it, so
x3+6x2+11x+6=(x+1)(x+2)(x+3)x^{3}+6x^{2}+11x+6=(x+1)(x+2)(x+3)
 Step 2: Decompose into partial fractions, 2(x+1)(x+2)(x+3)=Ax+1+Bx+2+Dx+3\dfrac{2}{(x+1)(x+2)(x+3)}=\dfrac{A}{x+1}+\dfrac{B}{x+2}+\dfrac{D}{x+3}. By the cover-up rule,
A=2(1)(2)=1,B=2(−1)(1)=−2,D=2(−2)(−1)=1A=\dfrac{2}{(1)(2)}=1,\qquad B=\dfrac{2}{(-1)(1)}=-2,\qquad D=\dfrac{2}{(-2)(-1)}=1
⇒2(x+1)(x+2)(x+3)=1x+1−2x+2+1x+3\Rightarrow \dfrac{2}{(x+1)(x+2)(x+3)}=\dfrac{1}{x+1}-\dfrac{2}{x+2}+\dfrac{1}{x+3}
 Step 3: Integrate and apply the limits:
∫01=[ln⁡(x+1)−2ln⁡(x+2)+ln⁡(x+3)]01\int_{0}^{1}=\Big[\ln(x+1)-2\ln(x+2)+\ln(x+3)\Big]_{0}^{1}
 Step 4: Evaluate. At x=1x=1: ln⁡2−2ln⁡3+ln⁡4=3ln⁡2−2ln⁡3\ln 2-2\ln 3+\ln 4=3\ln 2-2\ln 3. At x=0x=0: ln⁡1−2ln⁡2+ln⁡3=−2ln⁡2+ln⁡3\ln 1-2\ln 2+\ln 3=-2\ln 2+\ln 3.
I=(3ln⁡2−2ln⁡3)−(−2ln⁡2+ln⁡3)=5ln⁡2−3ln⁡3I=(3\ln 2-2\ln 3)-(-2\ln 2+\ln 3)=5\ln 2-3\ln 3
 Correct answer: (1)
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