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Exercise 4.2 · Q7

Q.Evaluate: ∫231x2+5x+6 dx\int_2^3 \dfrac{1}{x^2+5x+6}\,dx

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✓ Free question

x2+5x+6=(x+2)(x+3)x^2+5x+6=(x+2)(x+3), so 1(x+2)(x+3)=1x+2−1x+3\dfrac1{(x+2)(x+3)}=\dfrac1{x+2}-\dfrac1{x+3}.

∫23(1x+2−1x+3)dx=[ln⁡∣x+2∣−ln⁡∣x+3∣]23=(ln⁡5−ln⁡6)−(ln⁡4−ln⁡5)=2ln⁡5−ln⁡6−ln⁡4=ln⁡2524.\int_2^3\Big(\frac1{x+2}-\frac1{x+3}\Big)dx=\big[\ln|x+2|-\ln|x+3|\big]_2^3=(\ln5-\ln6)-(\ln4-\ln5)=2\ln5-\ln6-\ln4=\ln\frac{25}{24}.

✓Final answer

∫23dxx2+5x+6=ln⁡2524\displaystyle\int_2^3\frac{dx}{x^2+5x+6}=\ln\frac{25}{24}

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