Menu Close

find-arctan-x-x-dx-

Tinku Tara 0 Comments
Pin




Question Number 86955 by abdomathmax last updated on 01/Apr/20
find ∫ ((arctan(x))/x)dx
$${find}\:\int\:\:\frac{{arctan}\left({x}\right)}{{x}}{dx} \\ $$
Commented by Ar Brandon last updated on 01/Apr/20
Let f(a)=∫((arctan(ax))/x)dx ⇒f ′ (a)=∫(x/x)∙(1/(1+(ax)^2 ))dx=∫(1/(1+(ax)^2 ))dx ⇒f ′ (a) = (1/a)∙arctan(ax)+C .....
$${Let}\:{f}\left({a}\right)=\int\frac{{arctan}\left({ax}\right)}{{x}}{dx} \\ $$$$\Rightarrow{f}\:'\:\left({a}\right)=\int\frac{{x}}{{x}}\centerdot\frac{\mathrm{1}}{\mathrm{1}+\left({ax}\right)^{\mathrm{2}} }{dx}=\int\frac{\mathrm{1}}{\mathrm{1}+\left({ax}\right)^{\mathrm{2}} }{dx} \\ $$$$\Rightarrow{f}\:'\:\:\left({a}\right)\:=\:\frac{\mathrm{1}}{{a}}\centerdot{arctan}\left({ax}\right)+{C} \\ $$$$….. \\ $$
Answered by redmiiuser last updated on 01/Apr/20
arctan x=x−(x^3 /3)+(x^5 /5)−(x^7 /7)+... =Σ_(n=0) ^∞ (((−1)^n .x^(2n+1) )/(2n+1)) ∫Σ_(n=0) ^∞ (((−1)^n .x^(2n) .dx)/(2n+1)) =Σ_(n=0) ^∞ (((−1)^n .x^(2n+1) )/((2n+1)^2 ))
$${arc}\mathrm{tan}\:{x}={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}}−\frac{{x}^{\mathrm{7}} }{\mathrm{7}}+… \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} .{x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\int\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} .{x}^{\mathrm{2}{n}} .{dx}}{\mathrm{2}{n}+\mathrm{1}} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} .{x}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$
Commented by redmiiuser last updated on 01/Apr/20
Mr.abdomathmax pls check my answer and comment the appropriate.
$${Mr}.{abdomathmax}\:{pls} \\ $$$${check}\:{my}\:{answer}\:{and} \\ $$$${comment}\:{the}\:{appropriate}. \\ $$
Commented by mathmax by abdo last updated on 01/Apr/20
your answer is correct because the radius of arctan(x)is +∞ (integr serie)
$${your}\:{answer}\:{is}\:{correct}\:\:{because}\:{the}\:{radius}\:{of}\:{arctan}\left({x}\right){is}\:+\infty \\ $$$$\left({integr}\:{serie}\right) \\ $$
Answered by mind is power last updated on 02/Apr/20
=ln(x)arctan(x)−∫((ln(x))/(1+x^2 ))dx =ln(x)arctan(x)−∫((ln(x)dx)/((x+i)(x−i))) =ln(x)arctan(x)−(1/(2i))∫{((ln(x))/(x−i))−((ln(x))/(x+i))}dx =arctan(x)ln(x)−(1/(2i))∫((ln(x)dx)/(x−i))+(1/(2i))∫((ln(x))/(x+i))dx ∫((ln(x))/(x+a))dx=∫((ln(y−a))/y)dy=∫((ln(−a)+ln(1−(y/a)))/y)dy =ln(−a)ln(x+a)−Li_2 ((x/a)+1) =arctan(x)ln(x)−(1/(2i))ln(i)ln(x−i)+(1/(2i))Li_2 (ix+1)+(1/(2i))ln(−i)ln(x+i) −(1/(2i))Li_2 (−ix+1)+c
$$={ln}\left({x}\right){arctan}\left({x}\right)−\int\frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$={ln}\left({x}\right){arctan}\left({x}\right)−\int\frac{{ln}\left({x}\right){dx}}{\left({x}+{i}\right)\left({x}−{i}\right)} \\ $$$$={ln}\left({x}\right){arctan}\left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}{i}}\int\left\{\frac{{ln}\left({x}\right)}{{x}−{i}}−\frac{{ln}\left({x}\right)}{{x}+{i}}\right\}{dx} \\ $$$$={arctan}\left({x}\right){ln}\left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}{i}}\int\frac{{ln}\left({x}\right){dx}}{{x}−{i}}+\frac{\mathrm{1}}{\mathrm{2}{i}}\int\frac{{ln}\left({x}\right)}{{x}+{i}}{dx} \\ $$$$\int\frac{{ln}\left({x}\right)}{{x}+{a}}{dx}=\int\frac{{ln}\left({y}−{a}\right)}{{y}}{dy}=\int\frac{{ln}\left(−{a}\right)+{ln}\left(\mathrm{1}−\frac{{y}}{{a}}\right)}{{y}}{dy} \\ $$$$={ln}\left(−{a}\right){ln}\left({x}+{a}\right)−{Li}_{\mathrm{2}} \left(\frac{{x}}{{a}}+\mathrm{1}\right) \\ $$$$={arctan}\left({x}\right){ln}\left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left({i}\right){ln}\left({x}−{i}\right)+\frac{\mathrm{1}}{\mathrm{2}{i}}{Li}_{\mathrm{2}} \left({ix}+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(−{i}\right){ln}\left({x}+{i}\right) \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}{i}}{Li}_{\mathrm{2}} \left(−{ix}+\mathrm{1}\right)+{c} \\ $$$$ \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *