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Question Number 60036 by sitangshu17 last updated on 17/May/19

	Answered by tanmay last updated on 17/May/19
	$x^2 −5x+6>0 (x−2)(x−3)>0 f(x)=x^2 −5x+6 when i)x>3 [f(x)>0 ii)3>x>2 f(x)<0 iii)2>x f(x)>0 so x∈(−∞,2) ∪ (3,∞) now ∫lnxdx =lnx∫dx−∫[((d(lnx))/dx)∫dx]dx =xlnx−∫(1/x)×xdx =xlnx−x+c so ∫ln(x^2 −5x+6)dx ∫ln{(x−2)(x−3)}dx ∫ln(x−2)dx+∫ln(x−3)dx =(x−2)[ln(x−2)−1]+(x−3)[ln(x−3)−1]+c when x∈ (−∞,2)∪(3,∞)$
	$x^{2} - 5 x + 6 > 0$ $(x - 2) (x - 3) > 0$ $f (x) = x^{2} - 5 x + 6$ $w h e n$ $i) x > 3 [f (x) > 0$ $i i) 3 > x > 2 f (x) < 0$ $i i i) 2 > x f (x) > 0$ $s o x \in (- \infty, 2) \cup (3, \infty)$ $n o w$ $\int l n x d x$ $= l n x \int d x - \int [\frac{d (l n x)}{d x} \int d x] d x$ $= x l n x - \int \frac{1}{x} \times x d x$ $= x l n x - x + c$ $s o$ $\int l n (x^{2} - 5 x + 6) d x$ $\int l n {(x - 2) (x - 3)} d x$ $\int l n (x - 2) d x + \int l n (x - 3) d x$ $= (x - 2) [l n (x - 2) - 1] + (x - 3) [l n (x - 3) - 1] + c$ $w h e n x \in (- \infty, 2) \cup (3, \infty)$
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