Question Number 60036 by sitangshu17 last updated on 17/May/19
Answered by tanmay last updated on 17/May/19
![x^2 −5x+6>0 (x−2)(x−3)>0 f(x)=x^2 −5x+6 when i)x>3 [f(x)>0 ii)3>x>2 f(x)<0 iii)2>x f(x)>0 so x∈(−∞,2) ∪ (3,∞) now ∫lnxdx =lnx∫dx−∫[((d(lnx))/dx)∫dx]dx =xlnx−∫(1/x)×xdx =xlnx−x+c so ∫ln(x^2 −5x+6)dx ∫ln{(x−2)(x−3)}dx ∫ln(x−2)dx+∫ln(x−3)dx =(x−2)[ln(x−2)−1]+(x−3)[ln(x−3)−1]+c when x∈ (−∞,2)∪(3,∞)](https://www.tinkutara.com/question/Q60044.png)
$${x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6}>\mathrm{0} \\ $$$$\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)>\mathrm{0} \\ $$$${f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6} \\ $$$${when}\: \\ $$$$\left.{i}\right){x}>\mathrm{3}\:\:\:\:\:\left[{f}\left({x}\right)>\mathrm{0}\right. \\ $$$$\left.{ii}\right)\mathrm{3}>{x}>\mathrm{2}\:\:\:\:\:{f}\left({x}\right)<\mathrm{0} \\ $$$$\left.{iii}\right)\mathrm{2}>{x}\:\:\:{f}\left({x}\right)>\mathrm{0} \\ $$$${so}\:\:{x}\in\left(−\infty,\mathrm{2}\right)\:\cup\:\left(\mathrm{3},\infty\right) \\ $$$${now} \\ $$$$\int{lnxdx} \\ $$$$={lnx}\int{dx}−\int\left[\frac{{d}\left({lnx}\right)}{{dx}}\int{dx}\right]{dx} \\ $$$$={xlnx}−\int\frac{\mathrm{1}}{{x}}×{xdx} \\ $$$$={xlnx}−{x}+{c} \\ $$$${so} \\ $$$$\int{ln}\left({x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6}\right){dx} \\ $$$$\int{ln}\left\{\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)\right\}{dx} \\ $$$$\int{ln}\left({x}−\mathrm{2}\right){dx}+\int{ln}\left({x}−\mathrm{3}\right){dx} \\ $$$$=\left({x}−\mathrm{2}\right)\left[{ln}\left({x}−\mathrm{2}\right)−\mathrm{1}\right]+\left({x}−\mathrm{3}\right)\left[{ln}\left({x}−\mathrm{3}\right)−\mathrm{1}\right]+\boldsymbol{{c}} \\ $$$$\boldsymbol{{when}}\:\boldsymbol{{x}}\in\:\left(−\infty,\mathrm{2}\right)\cup\left(\mathrm{3},\infty\right) \\ $$