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Question Number 119282 Answers: 1 Comments: 0

... nice calculus... evaluate:: I:= ∫_0 ^( 1) li_2 (1−x^2 )dx=?? .m.n.1970.

$$\:\:\:\:\:\:\:\:...\:\:{nice}\:\:{calculus}... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{evaluate}:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{I}:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {li}_{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right){dx}=?? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.{m}.{n}.\mathrm{1970}. \\ $$$$\:\:\:\:\:\:\:\: \\ $$

Question Number 119318 Answers: 0 Comments: 0

find ∫_0 ^∞ ((lnx)/(x^4 +x^2 +2))dx

$${find}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{lnx}}{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{2}}{dx} \\ $$

Question Number 119317 Answers: 0 Comments: 0

calculate ∫_0 ^(2π) (dθ/((x^2 −2cosθ x+1)^2 ))

$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{d}\theta}{\left({x}^{\mathrm{2}} −\mathrm{2}{cos}\theta\:{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$

Question Number 119262 Answers: 3 Comments: 0

∫ (x^2 /( (√((4−x^2 )^5 )))) dx

$$\:\int\:\frac{{x}^{\mathrm{2}} }{\:\sqrt{\left(\mathrm{4}−{x}^{\mathrm{2}} \right)^{\mathrm{5}} }}\:{dx}\: \\ $$

Question Number 119256 Answers: 0 Comments: 0

Determine ∫_(−(π/4)) ^( (π/4)) (cost+(√(1+t^2 sin^3 tcos^3 t))dt=?

$$\: \\ $$$$\:\:\:\:\:\:{Determine}\:\:\int_{−\frac{\pi}{\mathrm{4}}} ^{\:\frac{\pi}{\mathrm{4}}} \left({cost}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} {sin}^{\mathrm{3}} {tcos}^{\mathrm{3}} {t}}{dt}=?\right. \\ $$

Question Number 119151 Answers: 0 Comments: 0

Question Number 119070 Answers: 2 Comments: 0

∫ ((x^2 −x+6)/(x^3 +3x)) dx ∫ ((5x^2 +3x−2)/(x^3 +2x^2 )) dx

$$\int\:\frac{{x}^{\mathrm{2}} −{x}+\mathrm{6}}{{x}^{\mathrm{3}} +\mathrm{3}{x}}\:{dx}\: \\ $$$$\int\:\frac{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{2}}{{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} }\:{dx}\: \\ $$

Question Number 119038 Answers: 2 Comments: 0

∫_(−π/4) ^(+π/4) ((√(1+tan x))/( (√(1−tan x))))dx

$$\underset{−\pi/\mathrm{4}} {\overset{+\pi/\mathrm{4}} {\int}}\frac{\sqrt{\mathrm{1}+\mathrm{tan}\:{x}}}{\:\sqrt{\mathrm{1}−\mathrm{tan}\:{x}}}{dx} \\ $$

Question Number 119021 Answers: 1 Comments: 0

Question Number 119006 Answers: 2 Comments: 0

Question Number 118997 Answers: 3 Comments: 0

Question Number 118959 Answers: 2 Comments: 0

∫ (dx/(x^6 −x^3 )) ?

$$\:\int\:\frac{{dx}}{{x}^{\mathrm{6}} −{x}^{\mathrm{3}} }\:? \\ $$

Question Number 118955 Answers: 1 Comments: 0

find ∫_0 ^∞ ((lnx)/(x^2 +x+1))dx

$${find}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{lnx}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx} \\ $$

Question Number 118949 Answers: 0 Comments: 0

find ∫_0 ^∞ ((sin(3cosx))/((x^2 +4)^2 ))dx

$${find}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left(\mathrm{3}{cosx}\right)}{\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }{dx} \\ $$

Question Number 118934 Answers: 0 Comments: 0

I=∫_0 ^∞ ((lnx)/(x^2 +2x+4)) dx put x+1=(√3) tanθ I=∫_0 ^(π/2) ((ln((√3) tanθ−1))/(3(sec^2 θ))) (√3) sec^2 θdθ I=(1/( (√3)))∫_0 ^(π/2) ln((√3)tanθ−1) dθ I=(1/( (√3)))∫_0 ^(π/2) ln((√3)sinθ−cosθ)−lncosθ dθ I=(1/( (√3)))∫_0 ^(π/2) ln(sin(θ−(π/6))+ln2−lncosθ dθ A=∫_(−2) ^6 ∣g(x)∣ dx ⇒let g(p)=0⇒ f(0)=p=2 A=∫_(−2) ^2 −g(x) dx+∫_2 ^6 g(x) dx ⇒put x=f(t) A=∫_(−1) ^0 −tf′(t) dt+∫_0 ^1 t f(t) dt A=∫_(−1) ^0 −3t^3 −3t dt+∫_0 ^1 3t^3 +3t dt A=2(((3t^4 )/4)+((3t^2 )/2))_0 ^1 =(9/2) lnx=(1/x)⇒xlnx=1(let x=α) ∫_1 ^α (1/x)−lnx dx=∫_α ^a lnx−(1/x) dx ⇒lnx−xlnx+x]_1 ^α =xlnx−x−lnx]_α ^a ⇒lnα−1+α−1=alna−a−lna−1+α +lnα ⇒alna−a−lna=−1 ⇒alna−lna=a−1⇒a=e s(t)=(1/2)∣OA^→ ×OB^→ ∣ s(t)=(1/2)∣(2,2,1)×(t,1,t+1)∣ s(t)=(1/2)∣(2t+1),(−2−t),(2−2t)∣ f(x)=(1/4)∫_0 ^x (2t+1)^2 +(t+2)^2 +(2−2t)^2 dt f(x)=(1/4)∫_0 ^x 9t^2 +9 dt=((3x^3 +9x)/4) A=(1/4)∫_0 ^6 (3x^3 +9x)dx=((3x^4 +18x^2 )/(16))]_0 ^6 A=((3.6^3 (6+1))/(16))=((567)/2)

$$ \\ $$$$ \\ $$$${I}=\int_{\mathrm{0}} ^{\infty} \:\frac{{lnx}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}}\:{dx} \\ $$$${put}\:{x}+\mathrm{1}=\sqrt{\mathrm{3}}\:{tan}\theta \\ $$$${I}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \:\frac{{ln}\left(\sqrt{\mathrm{3}}\:{tan}\theta−\mathrm{1}\right)}{\mathrm{3}\left({sec}^{\mathrm{2}} \theta\right)}\:\sqrt{\mathrm{3}}\:{sec}^{\mathrm{2}} \theta{d}\theta \\ $$$${I}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \:\:{ln}\left(\sqrt{\mathrm{3}}{tan}\theta−\mathrm{1}\right)\:{d}\theta \\ $$$${I}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {ln}\left(\sqrt{\mathrm{3}}{sin}\theta−{cos}\theta\right)−{lncos}\theta\:{d}\theta \\ $$$${I}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {ln}\left({sin}\left(\theta−\frac{\pi}{\mathrm{6}}\right)+{ln}\mathrm{2}−{lncos}\theta\:{d}\theta\right. \\ $$$$ \\ $$$$ \\ $$$${A}=\int_{−\mathrm{2}} ^{\mathrm{6}} \mid{g}\left({x}\right)\mid\:{dx} \\ $$$$\Rightarrow{let}\:{g}\left({p}\right)=\mathrm{0}\Rightarrow\:{f}\left(\mathrm{0}\right)={p}=\mathrm{2} \\ $$$${A}=\int_{−\mathrm{2}} ^{\mathrm{2}} −{g}\left({x}\right)\:{dx}+\int_{\mathrm{2}} ^{\mathrm{6}} {g}\left({x}\right)\:{dx} \\ $$$$\Rightarrow{put}\:{x}={f}\left({t}\right) \\ $$$${A}=\int_{−\mathrm{1}} ^{\mathrm{0}} −{tf}'\left({t}\right)\:{dt}+\int_{\mathrm{0}} ^{\mathrm{1}} {t}\:{f}\left({t}\right)\:{dt} \\ $$$${A}=\int_{−\mathrm{1}} ^{\mathrm{0}} −\mathrm{3}{t}^{\mathrm{3}} −\mathrm{3}{t}\:{dt}+\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{3}{t}^{\mathrm{3}} +\mathrm{3}{t}\:{dt} \\ $$$${A}=\mathrm{2}\left(\frac{\mathrm{3}{t}^{\mathrm{4}} }{\mathrm{4}}+\frac{\mathrm{3}{t}^{\mathrm{2}} }{\mathrm{2}}\right)_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{9}}{\mathrm{2}} \\ $$$${lnx}=\frac{\mathrm{1}}{{x}}\Rightarrow{xlnx}=\mathrm{1}\left({let}\:{x}=\alpha\right) \\ $$$$\int_{\mathrm{1}} ^{\alpha} \frac{\mathrm{1}}{{x}}−{lnx}\:{dx}=\int_{\alpha} ^{{a}} {lnx}−\frac{\mathrm{1}}{{x}}\:{dx} \\ $$$$\left.\Rightarrow\left.{lnx}−{xlnx}+{x}\right]_{\mathrm{1}} ^{\alpha} ={xlnx}−{x}−{lnx}\right]_{\alpha} ^{{a}} \\ $$$$\Rightarrow{ln}\alpha−\mathrm{1}+\alpha−\mathrm{1}={alna}−{a}−{lna}−\mathrm{1}+\alpha \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{ln}\alpha \\ $$$$\Rightarrow{alna}−{a}−{lna}=−\mathrm{1} \\ $$$$\Rightarrow{alna}−{lna}={a}−\mathrm{1}\Rightarrow{a}={e} \\ $$$$ \\ $$$$ \\ $$$${s}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mid{O}\overset{\rightarrow} {{A}}×{O}\overset{\rightarrow} {{B}}\mid \\ $$$${s}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mid\left(\mathrm{2},\mathrm{2},\mathrm{1}\right)×\left({t},\mathrm{1},{t}+\mathrm{1}\right)\mid \\ $$$${s}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mid\left(\mathrm{2}{t}+\mathrm{1}\right),\left(−\mathrm{2}−{t}\right),\left(\mathrm{2}−\mathrm{2}{t}\right)\mid \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{{x}} \left(\mathrm{2}{t}+\mathrm{1}\right)^{\mathrm{2}} +\left({t}+\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{2}−\mathrm{2}{t}\right)^{\mathrm{2}} \:{dt} \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{{x}} \mathrm{9}{t}^{\mathrm{2}} +\mathrm{9}\:{dt}=\frac{\mathrm{3}{x}^{\mathrm{3}} +\mathrm{9}{x}}{\mathrm{4}} \\ $$$$\left.{A}=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{6}} \left(\mathrm{3}{x}^{\mathrm{3}} +\mathrm{9}{x}\right){dx}=\frac{\mathrm{3}{x}^{\mathrm{4}} +\mathrm{18}{x}^{\mathrm{2}} }{\mathrm{16}}\right]_{\mathrm{0}} ^{\mathrm{6}} \\ $$$${A}=\frac{\mathrm{3}.\mathrm{6}^{\mathrm{3}} \left(\mathrm{6}+\mathrm{1}\right)}{\mathrm{16}}=\frac{\mathrm{567}}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 118928 Answers: 1 Comments: 2

Question Number 118927 Answers: 0 Comments: 0

Question Number 118924 Answers: 1 Comments: 0

... advanced calculus... prove that :: Σ_(n=1) ^∞ (((−1)^n H_n )/n^2 ) =∫_0 ^( 1) ((ln(1−x)ln(1+x) )/x)dx note :: H_n =Σ_(k=1) ^n (1/k) therefore: Σ_(n=1 ) ^∞ (((−1)^n H_n )/n^2 )=((−5)/8) ζ (3 ) ✓ ..m.n.july.1970...

$$\:\:\:\:\:\:\:\:\:...\:{advanced}\:{calculus}... \\ $$$$\:\:\:\:{prove}\:{that}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {H}_{{n}} }{{n}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}\right){ln}\left(\mathrm{1}+{x}\right)\:\:}{{x}}{dx}\:\: \\ $$$$\:\:\:\:\:{note}\:::\:{H}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}} \\ $$$$\:\:\:\:\:\:\:{therefore}:\:\underset{{n}=\mathrm{1}\:} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {H}_{{n}} }{{n}^{\mathrm{2}} }=\frac{−\mathrm{5}}{\mathrm{8}}\:\zeta\:\left(\mathrm{3}\:\right)\:\checkmark \\ $$$$\:\:\:\:\:\:\:\:..{m}.{n}.{july}.\mathrm{1970}... \\ $$$$ \\ $$

Question Number 118905 Answers: 4 Comments: 0

∫ (dλ/((λ^2 −9)^2 )) =?

$$\:\:\:\int\:\frac{{d}\lambda}{\left(\lambda^{\mathrm{2}} −\mathrm{9}\right)^{\mathrm{2}} }\:=?\: \\ $$

Question Number 118891 Answers: 2 Comments: 0

∫_0 ^π arctan(3^(cosx) )dx=??? please help

$$\:\int_{\mathrm{0}} ^{\pi} \boldsymbol{{arctan}}\left(\mathrm{3}^{\boldsymbol{{cosx}}} \right)\boldsymbol{{dx}}=??? \\ $$$$ \\ $$$$\boldsymbol{{please}}\:\boldsymbol{{help}} \\ $$

Question Number 118886 Answers: 1 Comments: 0

... advanced calculus... evaluate :: {_(2. Ω_2 = ∫_0 ^( (1/2)) ((ln^2 (1+x))/x) dx=??) ^(1. Ω_1 =∫_0 ^( (1/2)) ((ln^2 (1−x))/x)dx=??) ... M.N.1970...

$$\:\:\:\:\:\:\:\:\:\:...\:\:{advanced}\:\:{calculus}... \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:{evaluate}\:::\:\:\left\{_{\mathrm{2}.\:\Omega_{\mathrm{2}} =\:\int_{\mathrm{0}} ^{\:\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{{ln}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)}{{x}}\:{dx}=??} ^{\mathrm{1}.\:\Omega_{\mathrm{1}} =\int_{\mathrm{0}} ^{\:\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{{x}}{dx}=??} \right. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...\:\mathscr{M}.\mathscr{N}.\mathrm{1970}... \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 118834 Answers: 2 Comments: 0

∫ ((x^4 +1)/(x^5 +4x^3 )) dx