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....nice calculus...= Titu′s lemma:: for any positive numbers : a_1 ,a_2 ,...,a_n , b_1 ,b_2 ,...,b_n we have: (((a_1 +...+a_n )^2 )/(b_1 +...+b_n ))≤(a_1 ^2 /b_1 ) +...+(a_n ^2 /b_n ) proof : put : x=(x_1 ,...,x_n )∈R^n :y=(y_1 ,...,y_n )∈R^n (x.y)^2 ≤∣x∣^2 ∣y∣^2 (cauchy−schwarz inequality) (x_1 y_1 +...+x_n y_n )^2 ≤(x_(1 ) ^2 +...+x_n ^2 )(y_1 ^2 +...+y_(n ) ^2 ) by applying subsitution : x_i =(a_i /( (√b_i ))) , y_i =(√b_i ) (i=1,2 ,...,n) ((a_(1 ) ^2 +...+a_(n ) ^2 )/(b_2 +...+b_n ))≤(a_1 ^2 /b_1 )+...+(a_n ^2 /b_n ) ✓✓

$$\:\:\:\:\:\:\:\:\:\:\:\:\:....{nice}\:\:{calculus}...= \\ $$$$\:\:{Titu}'{s}\:{lemma}:: \\ $$$$\:{for}\:{any}\:{positive}\:{numbers}\:: \\ $$$${a}_{\mathrm{1}} ,{a}_{\mathrm{2}} ,...,{a}_{{n}} \:,\:{b}_{\mathrm{1}} ,{b}_{\mathrm{2}} ,...,{b}_{{n}} \\ $$$$\:{we}\:{have}: \\ $$$$\:\frac{\left({a}_{\mathrm{1}} +...+{a}_{{n}} \right)^{\mathrm{2}} }{{b}_{\mathrm{1}} +...+{b}_{{n}} }\leqslant\frac{{a}_{\mathrm{1}} ^{\mathrm{2}} }{{b}_{\mathrm{1}} }\:+...+\frac{{a}_{{n}} ^{\mathrm{2}} }{{b}_{{n}} }\: \\ $$$${proof}\:: \\ $$$${put}\::\:{x}=\left({x}_{\mathrm{1}} ,...,{x}_{{n}} \right)\in\mathbb{R}^{{n}} \\ $$$$\:\:\:\:\:\:\:\:\::{y}=\left({y}_{\mathrm{1}} ,...,{y}_{{n}} \right)\in\mathbb{R}^{{n}} \\ $$$$\left({x}.{y}\right)^{\mathrm{2}} \leqslant\mid{x}\mid^{\mathrm{2}} \mid{y}\mid^{\mathrm{2}} \left({cauchy}−{schwarz}\:\:{inequality}\right) \\ $$$$\left({x}_{\mathrm{1}} {y}_{\mathrm{1}} +...+{x}_{{n}} {y}_{{n}} \right)^{\mathrm{2}} \leqslant\left({x}_{\mathrm{1}\:\:\:} ^{\mathrm{2}} +...+{x}_{{n}} ^{\mathrm{2}} \right)\left({y}_{\mathrm{1}} ^{\mathrm{2}} +...+{y}_{{n}\:} ^{\mathrm{2}} \right) \\ $$$$\:{by}\:{applying}\:{subsitution}\:: \\ $$$$\:{x}_{{i}} =\frac{{a}_{{i}} }{\:\sqrt{{b}_{{i}} }}\:\:\:,\:\:{y}_{{i}} =\sqrt{{b}_{{i}} }\:\left({i}=\mathrm{1},\mathrm{2}\:,...,{n}\right) \\ $$$$\:\frac{{a}_{\mathrm{1}\:} ^{\mathrm{2}} +...+{a}_{{n}\:} ^{\mathrm{2}} }{{b}_{\mathrm{2}} +...+{b}_{{n}} }\leqslant\frac{{a}_{\mathrm{1}} ^{\mathrm{2}} }{{b}_{\mathrm{1}} }+...+\frac{{a}_{{n}} ^{\mathrm{2}} }{{b}_{{n}} }\:\:\:\checkmark\checkmark \\ $$$$ \\ $$

Question Number 127958 Answers: 1 Comments: 0

...nice calculus... calculate Ω=∫_1 ^( ∞) ((ln(x^4 −2x^2 +2))/(x(√(x^2 −1)) )) dx=?

$$\:\:\:\:\:\:\:\:\:\:\:...{nice}\:\:{calculus}... \\ $$$$\:\:\:{calculate} \\ $$$$\:\:\:\:\Omega=\int_{\mathrm{1}} ^{\:\infty} \frac{{ln}\left({x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}\right)}{{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:}\:{dx}=? \\ $$$$ \\ $$

Question Number 127948 Answers: 1 Comments: 0

Question Number 127952 Answers: 1 Comments: 0

Question Number 127925 Answers: 1 Comments: 0

find F(a)=∫_0 ^1 (√((1+a^2 t^2 )/(1−t^2 ))) dt for background see Q127811.

$${find}\:{F}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\frac{\mathrm{1}+{a}^{\mathrm{2}} {t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}} }}\:{dt} \\ $$$$ \\ $$$${for}\:{background}\:{see}\:{Q}\mathrm{127811}. \\ $$

Question Number 127904 Answers: 2 Comments: 0

prove that ∫_0 ^( 100) (dx/( (√(x(100−x))))) = π

$$\:\mathrm{prove}\:\mathrm{that}\:\int_{\mathrm{0}} ^{\:\mathrm{100}} \:\frac{\mathrm{dx}}{\:\sqrt{\mathrm{x}\left(\mathrm{100}−\mathrm{x}\right)}}\:=\:\pi \\ $$

Question Number 127885 Answers: 1 Comments: 0

∫(((sin (2tan^(−1) (x)+x))/x)) the limit [0,∞)

$$\int\left(\frac{\mathrm{sin}\:\left(\mathrm{2tan}^{−\mathrm{1}} \left({x}\right)+{x}\right)}{{x}}\right)\:\:{the}\:{limit}\:\left[\mathrm{0},\infty\right) \\ $$

Question Number 127870 Answers: 2 Comments: 0

2021 HAPPY NEW Year 1)∫((x^3 +3x+2)/((x^2 +1)^2 (x+1)))dx 2)∫((2cos(x)−sin(x))/(3sin(x)+5cos(x)))dx 3)∫((tan(2x))/( (√(sin^6 (x)+cos^6 (x)))))dx 4)∫x(√((1−x^2 )/(1+x^2 ))) dx

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2021} \\ $$$${HAPPY}\:{NEW}\:{Year} \\ $$$$\left.\mathrm{1}\right)\int\frac{{x}^{\mathrm{3}} +\mathrm{3}{x}+\mathrm{2}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \left({x}+\mathrm{1}\right)}{dx} \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\int\frac{\mathrm{2}{cos}\left({x}\right)−{sin}\left({x}\right)}{\mathrm{3}{sin}\left({x}\right)+\mathrm{5}{cos}\left({x}\right)}{dx} \\ $$$$ \\ $$$$\left.\mathrm{3}\right)\int\frac{{tan}\left(\mathrm{2}{x}\right)}{\:\sqrt{{sin}^{\mathrm{6}} \left({x}\right)+{cos}^{\mathrm{6}} \left({x}\right)}}{dx} \\ $$$$ \\ $$$$\left.\mathrm{4}\right)\int{x}\sqrt{\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }}\:{dx} \\ $$

Question Number 127857 Answers: 0 Comments: 1

∫(√x)e^x dx ?

$$\int\sqrt{{x}}{e}^{{x}} {dx}\:\:? \\ $$

Question Number 127851 Answers: 1 Comments: 0

ψ = ∫ (dx/(x^3 (((x^5 +1)^3 ))^(1/5) )) ?

$$\:\psi\:=\:\int\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{3}} \:\sqrt[{\mathrm{5}}]{\left(\mathrm{x}^{\mathrm{5}} +\mathrm{1}\right)^{\mathrm{3}} }}\:?\: \\ $$

Question Number 127833 Answers: 2 Comments: 0