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Question Number 90765 by john santu last updated on 26/Apr/20

if ∫ (ln(x))^2 dx =   x( ln^2 (x)+a ln(x)+b) +C  a,b , C are constant.   find the value of a and b

$${if}\:\int\:\left(\mathrm{ln}\left({x}\right)\right)^{\mathrm{2}} {dx}\:=\: \\ $$$${x}\left(\:\mathrm{ln}^{\mathrm{2}} \left({x}\right)+{a}\:\mathrm{ln}\left({x}\right)+{b}\right)\:+{C} \\ $$$${a},{b}\:,\:{C}\:{are}\:{constant}.\: \\ $$$${find}\:{the}\:{value}\:{of}\:{a}\:{and}\:{b}\: \\ $$

Commented by jagoll last updated on 26/Apr/20

∫ f(x) dx = h(x) ⇒ f(x) = h′(x)  ln^2 (x) = ln^2 (x)+a ln(x)+b +   x( ((2ln(x))/x)+ (a/x))   0 = (a+2)ln(x) + (a+b)    { ((a = −2)),((b = 2 )) :}

$$\int\:{f}\left({x}\right)\:{dx}\:=\:{h}\left({x}\right)\:\Rightarrow\:{f}\left({x}\right)\:=\:{h}'\left({x}\right) \\ $$$$\mathrm{ln}^{\mathrm{2}} \left({x}\right)\:=\:\mathrm{ln}^{\mathrm{2}} \left({x}\right)+{a}\:\mathrm{ln}\left({x}\right)+{b}\:+\: \\ $$$${x}\left(\:\frac{\mathrm{2ln}\left({x}\right)}{{x}}+\:\frac{{a}}{{x}}\right)\: \\ $$$$\mathrm{0}\:=\:\left({a}+\mathrm{2}\right)\mathrm{ln}\left({x}\right)\:+\:\left({a}+{b}\right)\: \\ $$$$\begin{cases}{{a}\:=\:−\mathrm{2}}\\{{b}\:=\:\mathrm{2}\:}\end{cases} \\ $$

Commented by john santu last updated on 26/Apr/20

good

$${good} \\ $$

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