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Question-94857




Question Number 94857 by i jagooll last updated on 21/May/20
Commented by prakash jain last updated on 21/May/20
Can you write without bold font?  It looks too dark.
$$\mathrm{Can}\:\mathrm{you}\:\mathrm{write}\:\mathrm{without}\:\mathrm{bold}\:\mathrm{font}? \\ $$$$\mathrm{It}\:\mathrm{looks}\:\mathrm{too}\:\mathrm{dark}. \\ $$
Commented by i jagooll last updated on 21/May/20
on my cellphone it is not dark
Commented by i jagooll last updated on 21/May/20
and that's the result of the photo
Commented by mr W last updated on 21/May/20
you should try to crop your image  properly, i.e. to keep more write  space around the formula (in this  case more write space on left and right  side from the formula), then you  will get properly sized image after  posting. currently the images you  posted look very tiny.  this is an example what i do:  tap your image and my image, you  will see where the difference is.
$${you}\:{should}\:{try}\:{to}\:{crop}\:{your}\:{image} \\ $$$${properly},\:{i}.{e}.\:{to}\:{keep}\:{more}\:{write} \\ $$$${space}\:{around}\:{the}\:{formula}\:\left({in}\:{this}\right. \\ $$$${case}\:{more}\:{write}\:{space}\:{on}\:{left}\:{and}\:{right} \\ $$$$\left.{side}\:{from}\:{the}\:{formula}\right),\:{then}\:{you} \\ $$$${will}\:{get}\:{properly}\:{sized}\:{image}\:{after} \\ $$$${posting}.\:{currently}\:{the}\:{images}\:{you} \\ $$$${posted}\:{look}\:{very}\:{tiny}. \\ $$$${this}\:{is}\:{an}\:{example}\:{what}\:{i}\:{do}: \\ $$$${tap}\:{your}\:{image}\:{and}\:{my}\:{image},\:{you} \\ $$$${will}\:{see}\:{where}\:{the}\:{difference}\:{is}. \\ $$
Commented by mr W last updated on 21/May/20
Commented by mr W last updated on 21/May/20
i think i know how the problem comes.  you write the formula with the editor  of this app and then save it as image  and then post it into the forum. but  the editor saves the formula as image  without any extra write space around  the formula. if you don′t treat such  an image with an other app, you  won′t be able to solve this problem.  but my question: why don′t you write  the formula directly in the forum?
$${i}\:{think}\:{i}\:{know}\:{how}\:{the}\:{problem}\:{comes}. \\ $$$${you}\:{write}\:{the}\:{formula}\:{with}\:{the}\:{editor} \\ $$$${of}\:{this}\:{app}\:{and}\:{then}\:{save}\:{it}\:{as}\:{image} \\ $$$${and}\:{then}\:{post}\:{it}\:{into}\:{the}\:{forum}.\:{but} \\ $$$${the}\:{editor}\:{saves}\:{the}\:{formula}\:{as}\:{image} \\ $$$${without}\:{any}\:{extra}\:{write}\:{space}\:{around} \\ $$$${the}\:{formula}.\:{if}\:{you}\:{don}'{t}\:{treat}\:{such} \\ $$$${an}\:{image}\:{with}\:{an}\:{other}\:{app},\:{you} \\ $$$${won}'{t}\:{be}\:{able}\:{to}\:{solve}\:{this}\:{problem}. \\ $$$${but}\:{my}\:{question}:\:{why}\:{don}'{t}\:{you}\:{write} \\ $$$${the}\:{formula}\:{directly}\:{in}\:{the}\:{forum}? \\ $$
Commented by Tinku Tara last updated on 21/May/20
We reverted changes so that images  will again be scaled to width. So  they will look big. Please update  to latest version. 2.71.  We are still working on a proper  solution for very small images.  Thank You.
$$\mathrm{We}\:\mathrm{reverted}\:\mathrm{changes}\:\mathrm{so}\:\mathrm{that}\:\mathrm{images} \\ $$$$\mathrm{will}\:\mathrm{again}\:\mathrm{be}\:\mathrm{scaled}\:\mathrm{to}\:\mathrm{width}.\:\mathrm{So} \\ $$$$\mathrm{they}\:\mathrm{will}\:\mathrm{look}\:\mathrm{big}.\:\mathrm{Please}\:\mathrm{update} \\ $$$$\mathrm{to}\:\mathrm{latest}\:\mathrm{version}.\:\mathrm{2}.\mathrm{71}. \\ $$$$\mathrm{We}\:\mathrm{are}\:\mathrm{still}\:\mathrm{working}\:\mathrm{on}\:\mathrm{a}\:\mathrm{proper} \\ $$$$\mathrm{solution}\:\mathrm{for}\:\mathrm{very}\:\mathrm{small}\:\mathrm{images}. \\ $$$$\mathrm{Thank}\:\mathrm{You}. \\ $$
Answered by mathmax by abdo last updated on 21/May/20
I =∫  (((x+1)^2 )/(x^2 (x^2 +1)^2 ))dx  complex method  let decompose  F(x) =((x^2  +2x+1)/(x^2 (x^2  +1)^2 )) ⇒ F(x) =((x^2  +2x+1)/(x^2 (x−i)^2 (x+i)^2 ))  F(x) =(a/x) +(b/x^2 ) +(c/(x−i)) +(d/((x−i)^2 )) +(e/(x+i)) +(f/((x+i)^2 ))  b =1   ,d =((2i)/(−(2i)^2 )) =((−2i)/(−4)) =(i/2)  f =((−2i)/(−(−2i)^2 )) =((2i)/(−4)) =−(i/2) ⇒F(x)=(a/x)+(1/x^2 ) +(c/(x−i)) +(i/(2(x−i)^2 ))  +(e/(x+i)) −(i/(2(x+i)^2 ))  lim_(x→+∞) xF(x) =0 =a +c +e  F(1) =1  =a +1 +(c/(1−i)) +(i/(2(1−i)^2 )) +(e/(1+i))−(i/(2(1+i)^2 ))  =a+1 +(1+i)(c/2) +(i/(2(−2i))) +(1−i)(e/2)−(i/(2(2i))) ⇒  a+(1/2) +(1/2)(1+i)c +(1/2)(1−i)e =1  after we calculate F(−1)  we get the value of a c ande ⇒  ∫ F(x)dx =aln∣x∣−(1/x) +cln(x−i)−(i/(2(x−i)))+eln(x+i)+(i/(2(x+i))) +C  =aln∣x∣ −(1/x) +cln(x−i) +eln(x+i)+(i/2)((1/(x+i))−(1/(x−i))) +C  =aln∣x∣ −(1/x) +cln(x−i)+eln(x+i) + (1/(x^2  +1)) +C
$$\mathrm{I}\:=\int\:\:\frac{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx}\:\:\mathrm{complex}\:\mathrm{method}\:\:\mathrm{let}\:\mathrm{decompose} \\ $$$$\mathrm{F}\left(\mathrm{x}\right)\:=\frac{\mathrm{x}^{\mathrm{2}} \:+\mathrm{2x}+\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\:\mathrm{F}\left(\mathrm{x}\right)\:=\frac{\mathrm{x}^{\mathrm{2}} \:+\mathrm{2x}+\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{x}−\mathrm{i}\right)^{\mathrm{2}} \left(\mathrm{x}+\mathrm{i}\right)^{\mathrm{2}} } \\ $$$$\mathrm{F}\left(\mathrm{x}\right)\:=\frac{\mathrm{a}}{\mathrm{x}}\:+\frac{\mathrm{b}}{\mathrm{x}^{\mathrm{2}} }\:+\frac{\mathrm{c}}{\mathrm{x}−\mathrm{i}}\:+\frac{\mathrm{d}}{\left(\mathrm{x}−\mathrm{i}\right)^{\mathrm{2}} }\:+\frac{\mathrm{e}}{\mathrm{x}+\mathrm{i}}\:+\frac{\mathrm{f}}{\left(\mathrm{x}+\mathrm{i}\right)^{\mathrm{2}} } \\ $$$$\mathrm{b}\:=\mathrm{1}\:\:\:,\mathrm{d}\:=\frac{\mathrm{2i}}{−\left(\mathrm{2i}\right)^{\mathrm{2}} }\:=\frac{−\mathrm{2i}}{−\mathrm{4}}\:=\frac{\mathrm{i}}{\mathrm{2}} \\ $$$$\mathrm{f}\:=\frac{−\mathrm{2i}}{−\left(−\mathrm{2i}\right)^{\mathrm{2}} }\:=\frac{\mathrm{2i}}{−\mathrm{4}}\:=−\frac{\mathrm{i}}{\mathrm{2}}\:\Rightarrow\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:+\frac{\mathrm{c}}{\mathrm{x}−\mathrm{i}}\:+\frac{\mathrm{i}}{\mathrm{2}\left(\mathrm{x}−\mathrm{i}\right)^{\mathrm{2}} } \\ $$$$+\frac{\mathrm{e}}{\mathrm{x}+\mathrm{i}}\:−\frac{\mathrm{i}}{\mathrm{2}\left(\mathrm{x}+\mathrm{i}\right)^{\mathrm{2}} } \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \mathrm{xF}\left(\mathrm{x}\right)\:=\mathrm{0}\:=\mathrm{a}\:+\mathrm{c}\:+\mathrm{e} \\ $$$$\mathrm{F}\left(\mathrm{1}\right)\:=\mathrm{1}\:\:=\mathrm{a}\:+\mathrm{1}\:+\frac{\mathrm{c}}{\mathrm{1}−\mathrm{i}}\:+\frac{\mathrm{i}}{\mathrm{2}\left(\mathrm{1}−\mathrm{i}\right)^{\mathrm{2}} }\:+\frac{\mathrm{e}}{\mathrm{1}+\mathrm{i}}−\frac{\mathrm{i}}{\mathrm{2}\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{2}} } \\ $$$$=\mathrm{a}+\mathrm{1}\:+\left(\mathrm{1}+\mathrm{i}\right)\frac{\mathrm{c}}{\mathrm{2}}\:+\frac{\mathrm{i}}{\mathrm{2}\left(−\mathrm{2i}\right)}\:+\left(\mathrm{1}−\mathrm{i}\right)\frac{\mathrm{e}}{\mathrm{2}}−\frac{\mathrm{i}}{\mathrm{2}\left(\mathrm{2i}\right)}\:\Rightarrow \\ $$$$\mathrm{a}+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{i}\right)\mathrm{c}\:+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{i}\right)\mathrm{e}\:=\mathrm{1} \\ $$$$\mathrm{after}\:\mathrm{we}\:\mathrm{calculate}\:\mathrm{F}\left(−\mathrm{1}\right)\:\:\mathrm{we}\:\mathrm{get}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{a}\:\mathrm{c}\:\mathrm{ande}\:\Rightarrow \\ $$$$\int\:\mathrm{F}\left(\mathrm{x}\right)\mathrm{dx}\:=\mathrm{aln}\mid\mathrm{x}\mid−\frac{\mathrm{1}}{\mathrm{x}}\:+\mathrm{cln}\left(\mathrm{x}−\mathrm{i}\right)−\frac{\mathrm{i}}{\mathrm{2}\left(\mathrm{x}−\mathrm{i}\right)}+\mathrm{eln}\left(\mathrm{x}+\mathrm{i}\right)+\frac{\mathrm{i}}{\mathrm{2}\left(\mathrm{x}+\mathrm{i}\right)}\:+\mathrm{C} \\ $$$$=\mathrm{aln}\mid\mathrm{x}\mid\:−\frac{\mathrm{1}}{\mathrm{x}}\:+\mathrm{cln}\left(\mathrm{x}−\mathrm{i}\right)\:+\mathrm{eln}\left(\mathrm{x}+\mathrm{i}\right)+\frac{\mathrm{i}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{x}+\mathrm{i}}−\frac{\mathrm{1}}{\mathrm{x}−\mathrm{i}}\right)\:+\mathrm{C} \\ $$$$=\mathrm{aln}\mid\mathrm{x}\mid\:−\frac{\mathrm{1}}{\mathrm{x}}\:+\mathrm{cln}\left(\mathrm{x}−\mathrm{i}\right)+\mathrm{eln}\left(\mathrm{x}+\mathrm{i}\right)\:+\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\:+\mathrm{C} \\ $$
Answered by niroj last updated on 21/May/20
 I= ∫  ((x^2 +2x+1)/({x(x^2 +1)}^2 ))dx     = ∫ ((x^2 +1+2x)/(x^2 (x^2 +1)^2 ))dx     =  ∫ (( (x^2 +1))/(x^2 (x^2 +1)^2 ))dx +∫ ((  2x)/(x^2 (x^2 +1)^2 ))dx   = ∫ (1/(x^2 (x^2 +1)))dx+∫ (( 2x)/(x^2 (x^2 +1)))dx    I=I_1 +I_2     For, I_2      Put,  x^2 +1=t , x^2 =t−1          2xdx=dt    ∫(( 1)/x^2 )dx−∫(( 1)/(x^2 +1))dx+∫(1/((t−1)t))dt     [ (x^(−2+1) /(−2+1))] −tan^(−1) x +∫(1/(t−1))dt−∫(1/t)dt+ C     (x^(−1) /(−1)) − tan^(−1) x +log (t−1)−log t+C  −(1/x) − tan^(−1) x +log ((t−1)/t) +C   − (1/x)−tan^(−1) x + log ((x^2 +1−1)/(x^2 +1))+C    −(1/x) −tan^(−1) x +log (x^2 /(x^2 +1))+C //.
$$\:\mathrm{I}=\:\int\:\:\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{1}}{\left\{\mathrm{x}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\right\}^{\mathrm{2}} }\mathrm{dx} \\ $$$$\:\:\:=\:\int\:\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{1}+\mathrm{2x}}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx} \\ $$$$\:\:\:=\:\:\int\:\frac{\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx}\:+\int\:\frac{\:\:\mathrm{2x}}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx} \\ $$$$\:=\:\int\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)}\mathrm{dx}+\int\:\frac{\:\mathrm{2x}}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)}\mathrm{dx} \\ $$$$\:\:\mathrm{I}={I}_{\mathrm{1}} +{I}_{\mathrm{2}} \\ $$$$\:\:\mathrm{For},\:{I}_{\mathrm{2}} \\ $$$$\:\:\:\mathrm{Put},\:\:\mathrm{x}^{\mathrm{2}} +\mathrm{1}=\mathrm{t}\:,\:\mathrm{x}^{\mathrm{2}} =\mathrm{t}−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{2xdx}=\mathrm{dt} \\ $$$$\:\:\int\frac{\:\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\mathrm{dx}−\int\frac{\:\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx}+\int\frac{\mathrm{1}}{\left(\mathrm{t}−\mathrm{1}\right)\mathrm{t}}\mathrm{dt} \\ $$$$\:\:\:\left[\:\frac{\mathrm{x}^{−\mathrm{2}+\mathrm{1}} }{−\mathrm{2}+\mathrm{1}}\right]\:−\mathrm{tan}^{−\mathrm{1}} \mathrm{x}\:+\int\frac{\mathrm{1}}{\mathrm{t}−\mathrm{1}}\mathrm{dt}−\int\frac{\mathrm{1}}{\mathrm{t}}\mathrm{dt}+\:\mathrm{C} \\ $$$$\:\:\:\frac{\mathrm{x}^{−\mathrm{1}} }{−\mathrm{1}}\:−\:\mathrm{tan}^{−\mathrm{1}} \mathrm{x}\:+\mathrm{log}\:\left(\mathrm{t}−\mathrm{1}\right)−\mathrm{log}\:\mathrm{t}+\mathrm{C} \\ $$$$−\frac{\mathrm{1}}{\mathrm{x}}\:−\:\mathrm{tan}^{−\mathrm{1}} \mathrm{x}\:+\mathrm{log}\:\frac{\mathrm{t}−\mathrm{1}}{\mathrm{t}}\:+\mathrm{C} \\ $$$$\:−\:\frac{\mathrm{1}}{\mathrm{x}}−\mathrm{tan}^{−\mathrm{1}} \mathrm{x}\:+\:\mathrm{log}\:\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}+\mathrm{C} \\ $$$$\:\:−\frac{\mathrm{1}}{\mathrm{x}}\:−\mathrm{tan}^{−\mathrm{1}} \mathrm{x}\:+\mathrm{log}\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}+\mathrm{C}\://. \\ $$$$\: \\ $$$$ \\ $$

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