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∫12x dx(x+1)(x+2)=\int_1^2 \frac{x \, dx}{(x+1)(x+2)} =

Solution

✅ Correct Option: 3

Partial fractions: x(x+1)(x+2)=−1x+1+2x+2\frac{x}{(x+1)(x+2)}=\frac{-1}{x+1}+\frac{2}{x+2}.

∫12[−1x+1+2x+2]dx=[−ln⁡∣x+1∣+2ln⁡∣x+2∣]12\int_1^2\left[\frac{-1}{x+1}+\frac{2}{x+2}\right]dx=[-\ln|x+1|+2\ln|x+2|]_1^2

=(−ln⁡3+2ln⁡4)−(−ln⁡2+2ln⁡3)=ln⁡3227=(-\ln 3+2\ln 4)-(-\ln 2+2\ln 3)=\ln\frac{32}{27}.

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