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Question Number 67687    Answers: 0   Comments: 3

find the range of values of ∣((x^2 −9)/3_ )∣= ((9−x^2 )/3)

$${find}\:{the}\:{range}\:{of}\:{values}\:{of}\: \\ $$$$\:\:\mid\frac{{x}^{\mathrm{2}} −\mathrm{9}}{\mathrm{3}_{\:} }\mid=\:\frac{\mathrm{9}−{x}^{\mathrm{2}} }{\mathrm{3}} \\ $$$$ \\ $$

Question Number 67686    Answers: 0   Comments: 2

given that the roots of the equation 4x^2 + 6x + 9 =0 are λ and δ where λ = (1 + α^2 +β^2 ) and δ = α^3 + β^3 find an equation whose roots are (1/(αλ)) and (1/(βδ))

$${given}\:{that}\:{the}\:{roots}\:{of}\:{the}\:{equation}\:\:\mathrm{4}{x}^{\mathrm{2}} \:+\:\mathrm{6}{x}\:+\:\mathrm{9}\:=\mathrm{0}\:{are}\:\:\lambda\:{and}\:\delta\:\:{where}\: \\ $$$$\:\lambda\:=\:\left(\mathrm{1}\:+\:\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} \right)\:\:{and}\:\:\delta\:=\:\alpha^{\mathrm{3}} \:+\:\beta^{\mathrm{3}} \\ $$$${find}\:{an}\:{equation}\:{whose}\:{roots}\:{are}\: \\ $$$$\:\:\frac{\mathrm{1}}{\alpha\lambda}\:{and}\:\:\frac{\mathrm{1}}{\beta\delta} \\ $$

Question Number 67684    Answers: 0   Comments: 2

given the function f(x) = { ((x^2 , for 0≤ x< 2)),((ax + 3, for 2≤ x < 4)) :} is periodic of period 4, and is continuous. a) Find the value of a. b) Find the valu of f(6) c) sketch the graph for y =f(x). help me please, for the graph i don′t know wbere to put y=x^2 and y = ax + 3 and where do i put a closed dot and an open dot.

$${given}\:{the}\:{function}\: \\ $$$${f}\left({x}\right)\:=\begin{cases}{{x}^{\mathrm{2}} \:\:,\:{for}\:\:\:\mathrm{0}\leqslant\:{x}<\:\mathrm{2}}\\{{ax}\:+\:\mathrm{3},\:{for}\:\:\mathrm{2}\leqslant\:{x}\:<\:\mathrm{4}}\end{cases} \\ $$$${is}\:{periodic}\:{of}\:{period}\:\:\mathrm{4},\:{and}\:{is}\:{continuous}. \\ $$$$\left.{a}\right)\:{Find}\:\:{the}\:{value}\:{of}\:\:{a}. \\ $$$$\left.{b}\right)\:{Find}\:{the}\:{valu}\:{of}\:\:{f}\left(\mathrm{6}\right) \\ $$$$\left.{c}\right)\:{sketch}\:{the}\:{graph}\:{for}\:{y}\:={f}\left({x}\right). \\ $$$${help}\:{me}\:{please},\:{for}\:{the}\:{graph}\:{i}\:{don}'{t}\:{know}\:{wbere}\:{to}\:{put}\:\:{y}={x}^{\mathrm{2}} \:{and}\:{y}\:=\:{ax}\:+\:\mathrm{3}\:{and} \\ $$$${where}\:{do}\:{i}\:{put}\:{a}\:{closed}\:\:{dot}\:{and}\:{an}\:{open}\:{dot}. \\ $$$$ \\ $$

Question Number 67674    Answers: 0   Comments: 3

let f(a) =∫_0 ^∞ (dx/((x^2 +1)(x^2 +a))) with a>0 1) determine a explicit form of f(a) 2) calculate g(a) =∫_0 ^∞ (dx/((x^2 +1)(x^2 +a)^2 )) 3)give f^((n)) (a) at form of integral 4)calculate ∫_0 ^∞ (dx/((x^2 +1)(x^2 +3)^2 )) and ∫_0 ^∞ (dx/((x^2 +1)^3 ))

$${let}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +{a}\right)}\:\:{with}\:{a}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{determine}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({a}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{g}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+{a}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{3}\right){give}\:{f}^{\left({n}\right)} \left({a}\right)\:{at}\:{form}\:{of}\:{integral} \\ $$$$\left.\mathrm{4}\right){calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} }\:{and} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} } \\ $$

Question Number 67673    Answers: 0   Comments: 1

calculate ∫_(−∞) ^(+∞) (dx/((x^2 −x+1)(x^2 +x+1)))

$${calculate}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)} \\ $$

Question Number 67672    Answers: 0   Comments: 2

decompose the folowing fraction at R(x) 1)F(x)=(x^3 /(1−x^6 )) 2) G(x) =((x^2 +1)/(x^3 (x^2 +x+1)^2 ))

$${decompose}\:{the}\:{folowing}\:\:{fraction}\:{at}\:{R}\left({x}\right) \\ $$$$\left.\mathrm{1}\right){F}\left({x}\right)=\frac{{x}^{\mathrm{3}} }{\mathrm{1}−{x}^{\mathrm{6}} } \\ $$$$\left.\mathrm{2}\right)\:{G}\left({x}\right)\:=\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{3}} \left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$

Question Number 67664    Answers: 0   Comments: 2

show that ∃ n ∈ N^(+ ) : sin^n x + cos^n x = 1 and cosh^n x − sinh^n x = 1. Hint: use Induction method.

$${show}\:{that}\:\:\exists\:{n}\:\in\:{N}^{+\:} \::\:\:{sin}^{{n}} {x}\:+\:{cos}^{{n}} {x}\:=\:\mathrm{1}\:{and}\:\:{cosh}^{{n}} {x}\:−\:{sinh}^{{n}} {x}\:=\:\mathrm{1}. \\ $$$$ \\ $$$${Hint}:\:{use}\:{Induction}\:{method}. \\ $$$$ \\ $$

Question Number 67662    Answers: 0   Comments: 2

please explain the fact that ∫(1/x)dx = ln x + k

$${please}\:{explain}\:{the}\:{fact}\:{that}\: \\ $$$$\int\frac{\mathrm{1}}{{x}}{dx}\:=\:{ln}\:{x}\:+\:{k} \\ $$

Question Number 67660    Answers: 1   Comments: 3

Question Number 67659    Answers: 0   Comments: 3

Help me obtain the value of e from (1 + (1/n))^n how do i go about it.

$${Help}\:{me}\:{obtain}\:\:\:{the}\:{value}\:{of}\:\:{e}\:{from} \\ $$$$\:\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{{n}}\right)^{{n}} \:\:{how}\:{do}\:{i}\:{go}\:{about}\:{it}. \\ $$

Question Number 67655    Answers: 0   Comments: 3

Question Number 67653    Answers: 1   Comments: 0

Question Number 67651    Answers: 0   Comments: 0

1)Let consider S= Σ_(n=0) ^∞ n and T=Σ_(n=0) ^∞ (−1)^(n+1) n We know that ∀ x∈]−1;1] Σ_(n=0) ^∞ (−x)^n =(1/(1+x)) , then after derivating (1/((1+x)^2 ))=Σ_(n=1) ^∞ (−1)^(n+1) nx^(n−1) for x=1 ,we get T=(1/4) Now let ascertain something T=Σ_(n=0) ^∞ (−1)^(2n+2) (2n+1) +Σ_(n=0) ^∞ (−1)^(2n+1) (2n) =Σ_(n=0) ^∞ (2n+1) −2S (•) knowing that S=Σ_(n=0) ^∞ (2n)+Σ_(n=0) ^∞ (2n+1) So Σ_(n=0) ^∞ (2n+1)= S−2S When replacing that value in (•) we get T=(S−2S )−2S If i conclude that T=−3S and finally find S=−(T/3)=−(1/(12)) where will the mistake be? 2)Let consider K=1+((2020)/(2019))+(((2020)/(2019)))^2 +(((2020)/(2019)))^3 +......+... ((2020)/(2019))K=((2020)/(2019))+(((2020)/(2019)))^2 +(((2020)/(2019)))^3 +.... K−1=((2020)/(2019))+(((2020)/(2019)))^2 +(((2020)/(2019)))^3 +... So ((2020)/(2019))K=K−1 Then K=−2019 Where is the error? 3)Let consider n an integer We have 0=n−n =n+(−n)=n+(−n)^(2×(1/2)) =n+[(−n)^2 ]^(1/2) =n+ (√n^2 ) = n+n=2n So << all integer are null : 0 is the only integer>> Where is the error? 4) let consider n an integer different of zero and f(n)=nln(n) we have (df/dn)=ln(n)+1 (•) Likewise f(n)=ln(n^n ) and we know that n^n =n×n×n×......×n (n times) So f(n)=ln(n)+ln(n)+......+ln(n) (n times) Now we have (df/dn)=(1/n)+(1/n)+....+(1/n) (n times) So (df/( dn))=1 (••) Relation (•) and (••) give ln(n)+1=1 then ln(n)=0 ⇒ n=1 << The logarithm of all n≥1 is null : There is no integer big than 1 >> Where is the error? 5) let consider x=0,999999999....... we ascertain that 10x=9,999999999...... then 10x=9+0,999999999.... So 10x=9+x finally x=1 << 0.9999999999999999 ..... is and integer >> Is there any error? 6)let consider a=((26666666666666666)/(66666666666666665)) b=((999999999999999999999995)/(199999999999999999999999)) In the way to cancel , if i just remove one common figure to the numerator and to the denominator And i find a=(2/5) and b=(5/1) Will it be wrong? if no ,explain!

$$\left.\mathrm{1}\right){Let}\:{consider}\:\:{S}=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:{n}\:\:\:\:\:{and}\:\:{T}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {n} \\ $$$$\left.{W}\left.{e}\:{know}\:{that}\:\:\forall\:{x}\in\right]−\mathrm{1};\mathrm{1}\right] \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−{x}\right)^{{n}} \:=\frac{\mathrm{1}}{\mathrm{1}+{x}}\:,\:\: \\ $$$${then}\:{after}\:{derivating}\: \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \:{nx}^{{n}−\mathrm{1}} \\ $$$${for}\:\:{x}=\mathrm{1}\:,{we}\:{get}\:\:\:{T}=\frac{\mathrm{1}}{\mathrm{4}}\: \\ $$$${Now}\:\:{let}\:{ascertain}\:{something} \\ $$$${T}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{2}} \left(\mathrm{2}{n}+\mathrm{1}\right)\:+\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} \left(\mathrm{2}{n}\right) \\ $$$$\:\:=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\mathrm{2}{n}+\mathrm{1}\right)\:−\mathrm{2}{S}\:\:\:\:\:\:\:\:\:\:\:\:\left(\bullet\right) \\ $$$${knowing}\:{that}\:{S}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\mathrm{2}{n}\right)+\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\mathrm{2}{n}+\mathrm{1}\right) \\ $$$${So}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\mathrm{2}{n}+\mathrm{1}\right)=\:{S}−\mathrm{2}{S}\: \\ $$$${When}\:{replacing}\:{that}\:{value}\:{in}\:\:\left(\bullet\right) \\ $$$${we}\:{get}\:\: \\ $$$${T}=\left({S}−\mathrm{2}{S}\:\right)−\mathrm{2}{S} \\ $$$$\: \\ $$$${If}\:{i}\:{conclude}\:{that}\:\:{T}=−\mathrm{3}{S}\:\: \\ $$$${and}\:{finally}\:\:{find}\:\:{S}=−\frac{{T}}{\mathrm{3}}=−\frac{\mathrm{1}}{\mathrm{12}}\:\: \\ $$$${where}\:{will}\:{the}\:{mistake}\:{be}? \\ $$$$\left.\:\mathrm{2}\right){Let}\:{consider}\:{K}=\mathrm{1}+\frac{\mathrm{2020}}{\mathrm{2019}}+\left(\frac{\mathrm{2020}}{\mathrm{2019}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2020}}{\mathrm{2019}}\right)^{\mathrm{3}} +......+... \\ $$$$\frac{\mathrm{2020}}{\mathrm{2019}}{K}=\frac{\mathrm{2020}}{\mathrm{2019}}+\left(\frac{\mathrm{2020}}{\mathrm{2019}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2020}}{\mathrm{2019}}\right)^{\mathrm{3}} +.... \\ $$$${K}−\mathrm{1}=\frac{\mathrm{2020}}{\mathrm{2019}}+\left(\frac{\mathrm{2020}}{\mathrm{2019}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2020}}{\mathrm{2019}}\right)^{\mathrm{3}} +... \\ $$$${So}\:\:\frac{\mathrm{2020}}{\mathrm{2019}}{K}={K}−\mathrm{1}\: \\ $$$${Then}\:\:{K}=−\mathrm{2019} \\ $$$${Where}\:{is}\:{the}\:{error}? \\ $$$$\left.\mathrm{3}\right){Let}\:{consider}\:{n}\:{an}\:\:{integer}\: \\ $$$${We}\:{have} \\ $$$$\mathrm{0}={n}−{n} \\ $$$$\:\:={n}+\left(−{n}\right)={n}+\left(−{n}\right)^{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\:={n}+\left[\left(−{n}\right)^{\mathrm{2}} \right]^{\frac{\mathrm{1}}{\mathrm{2}}} ={n}+\:\sqrt{{n}^{\mathrm{2}} }\:=\:{n}+{n}=\mathrm{2}{n} \\ $$$${So}\:<<\:{all}\:\:{integer}\:\:\:{are}\:{null}\:\::\:\:\mathrm{0}\:{is}\:{the}\:\:{only}\:{integer}>> \\ $$$${Where}\:{is}\:{the}\:{error}? \\ $$$$\left.\mathrm{4}\right)\:\:\:{let}\:{consider}\:{n}\:{an}\:{integer}\:{different}\:{of}\:{zero}\:{and}\:\:{f}\left({n}\right)={nln}\left({n}\right) \\ $$$${we}\:{have}\:\:\frac{{df}}{{dn}}={ln}\left({n}\right)+\mathrm{1}\:\:\:\:\left(\bullet\right) \\ $$$${Likewise}\:{f}\left({n}\right)={ln}\left({n}^{{n}} \right)\:\:\:{and}\:{we}\:{know}\:{that} \\ $$$${n}^{{n}} ={n}×{n}×{n}×......×{n}\:\:\left({n}\:{times}\right) \\ $$$${So}\:\:{f}\left({n}\right)={ln}\left({n}\right)+{ln}\left({n}\right)+......+{ln}\left({n}\right)\:\:\:\:\:\:\left({n}\:{times}\right) \\ $$$${Now}\:{we}\:{have}\:\:\frac{{df}}{{dn}}=\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}}+....+\frac{\mathrm{1}}{{n}}\:\:\:\:\:\:\left({n}\:\:{times}\right) \\ $$$${So}\:\:\:\frac{{df}}{\:{dn}}=\mathrm{1}\:\:\:\:\:\left(\bullet\bullet\right) \\ $$$${Relation}\:\left(\bullet\right)\:\:{and}\:\:\:\:\left(\bullet\bullet\right)\:\:{give}\:\:\:\: \\ $$$${ln}\left({n}\right)+\mathrm{1}=\mathrm{1}\:\:\:{then}\:\:{ln}\left({n}\right)=\mathrm{0}\:\Rightarrow\:{n}=\mathrm{1} \\ $$$$<<\:{The}\:{logarithm}\:{of}\:\:{all}\:{n}\geqslant\mathrm{1}\:{is}\:{null}\::\:{There}\:{is}\:{no}\:{integer}\:{big}\:{than}\:\mathrm{1}\:>> \\ $$$${Where}\:{is}\:{the}\:{error}? \\ $$$$\left.\mathrm{5}\right)\:{let}\:{consider}\:\:{x}=\mathrm{0},\mathrm{999999999}....... \\ $$$${we}\:{ascertain}\:{that}\: \\ $$$$\mathrm{10}{x}=\mathrm{9},\mathrm{999999999}...... \\ $$$${then}\:\:\mathrm{10}{x}=\mathrm{9}+\mathrm{0},\mathrm{999999999}.... \\ $$$${So}\:\:\mathrm{10}{x}=\mathrm{9}+{x}\:\: \\ $$$${finally}\:\:{x}=\mathrm{1} \\ $$$$<<\:\mathrm{0}.\mathrm{9999999999999999}\:.....\:\:{is}\:\:{and}\:{integer}\:>> \\ $$$${Is}\:{there}\:{any}\:{error}? \\ $$$$\left.\mathrm{6}\right){let}\:{consider}\:\:{a}=\frac{\mathrm{26666666666666666}}{\mathrm{66666666666666665}}\: \\ $$$${b}=\frac{\mathrm{999999999999999999999995}}{\mathrm{199999999999999999999999}}\: \\ $$$${In}\:{the}\:{way}\:{to}\:{cancel}\:,\:{if}\:{i}\:{just}\:{remove}\:{one}\:{common} \\ $$$${figure}\:{to}\:{the}\:{numerator}\:{and}\:{to}\:{the}\:{denominator} \\ $$$${And}\:{i}\:{find}\:{a}=\frac{\mathrm{2}}{\mathrm{5}}\:\:{and}\:\:{b}=\frac{\mathrm{5}}{\mathrm{1}}\: \\ $$$${Will}\:{it}\:{be}\:{wrong}?\:{if}\:\:{no}\:,{explain}! \\ $$$$\: \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\: \\ $$

Question Number 67647    Answers: 0   Comments: 0

Question Number 67631    Answers: 0   Comments: 3

Can you please tell me, where does this formula come from? And what means the factorial of a non- integer number? π = ((1/2)!)^2 × 4 I′ve verified the above equation with calculator. Thank you

$$ \\ $$$$ \\ $$$$\:\:\:\mathrm{Can}\:\mathrm{you}\:\mathrm{please}\:\mathrm{tell}\:\mathrm{me},\:\mathrm{where}\:\mathrm{does}\:\mathrm{this}\: \\ $$$$\:\:\:\mathrm{formula}\:\mathrm{come}\:\mathrm{from}? \\ $$$$\:\:\:\mathrm{And}\:\mathrm{what}\:\mathrm{means}\:\mathrm{the}\:\mathrm{factorial}\:\mathrm{of}\:\mathrm{a}\:\mathrm{non}- \\ $$$$\:\:\:\mathrm{integer}\:\mathrm{number}? \\ $$$$ \\ $$$$\:\:\:\:\:\:\pi\:=\:\left(\frac{\mathrm{1}}{\mathrm{2}}!\right)^{\mathrm{2}} ×\:\mathrm{4} \\ $$$$ \\ $$$$\:\:\:{I}'{ve}\:{verified}\:{the}\:{above}\:{equation}\:{with}\:{calculator}. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Thank}\:\mathrm{you} \\ $$$$ \\ $$$$ \\ $$

Question Number 67628    Answers: 0   Comments: 0

∫x^n lnx/n^(x ) dx

$$\int{x}^{{n}} {lnx}/{n}^{{x}\:} \:{dx} \\ $$

Question Number 67623    Answers: 1   Comments: 5

Question Number 67618    Answers: 1   Comments: 0

find the area abovnded r=cos2θ

$${find}\:{the}\:{area}\:{abovnded}\:{r}={cos}\mathrm{2}\theta \\ $$

Question Number 67617    Answers: 0   Comments: 2

Question Number 67615    Answers: 2   Comments: 1

Question Number 67574    Answers: 0   Comments: 1

Solve x^2 +1<−5

$$\mathrm{Solve}\:\mathrm{x}^{\mathrm{2}} +\mathrm{1}<−\mathrm{5} \\ $$

Question Number 67572    Answers: 0   Comments: 1

∫_(−(π/2)) ^(π/2) {sin∣x∣+cos∣x∣} dx

$$\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \left\{{sin}\mid{x}\mid+{cos}\mid{x}\mid\right\}\:{dx} \\ $$

Question Number 67564    Answers: 0   Comments: 8

Question Number 67561    Answers: 1   Comments: 0

Question Number 67559    Answers: 0   Comments: 3

calculate Σ_(n=0) ^∞ (1/(n^(2 ) +1)) and Σ_(n=0) ^∞ (((−1)^n )/(n^2 +1))

$${calculate}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}\:} +\mathrm{1}}\:\:{and}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} \:+\mathrm{1}} \\ $$

Question Number 67542    Answers: 0   Comments: 4

let f(a) =∫_(−∞) ^(+∞) (dx/((x^2 +1)(a +e^(ix) ))) with a>0 1)find a explicit form of f(a) 2) determine also g(a)=∫_(−∞) ^(+∞) (dx/((x^2 +1)(a+e^(ix) )^2 )) 3)let I =Re(∫_(−∞) ^(+∞) (dx/((x^2 +1)(2+e^(ix) )))) and J=Im(∫_(−∞) ^(+∞) (dx/((x^2 +1)(2+e^x )))) determine I and J and its values.

$${let}\:{f}\left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({a}\:+{e}^{{ix}} \right)}\:\:\:{with}\:{a}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right){find}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({a}\right) \\ $$$$\left.\mathrm{2}\right)\:{determine}\:{also}\:{g}\left({a}\right)=\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({a}+{e}^{{ix}} \right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{3}\right){let}\:{I}\:={Re}\left(\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{2}+{e}^{{ix}} \right)}\right)\:{and}\:{J}={Im}\left(\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{2}+{e}^{{x}} \right)}\right) \\ $$$$\:{determine}\:{I}\:{and}\:{J}\:\:{and}\:\:{its}\:{values}. \\ $$

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