Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 124277 by SOMEDAVONG last updated on 02/Dec/20

I.∫((4x^2 lnx)/(1+x^4 ))dx=??

$$\mathrm{I}.\int\frac{\mathrm{4x}^{\mathrm{2}} \mathrm{lnx}}{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\mathrm{dx}=?? \\ $$

Commented by MJS_new last updated on 02/Dec/20

((4x^2 )/(x^4 +1))=((((√2)/2)(1−i))/(x−((√2)/2)(1+i)))+((((√2)/2)(1+i))/(x−((√2)/2)(1−i)))−((((√2)/2)(1+i))/(x+((√2)/2)(1−i)))−((((√2)/2)(1−i))/(x+((√2)/2)(1+i))) ⇒ we have to solve a∫((ln x)/(x−b))dx and then insert I get a∫((ln x)/(x−b))dx=a(ln (−b) ln ∣x+b∣ −Li_2 ((x/b)+1))+C but I′m not sure if this is right for a, b ∉R

$$\frac{\mathrm{4}{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} +\mathrm{1}}=\frac{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{i}\right)}{{x}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{i}\right)}+\frac{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{i}\right)}{{x}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{i}\right)}−\frac{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{i}\right)}{{x}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{i}\right)}−\frac{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{i}\right)}{{x}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{i}\right)} \\ $$$$\Rightarrow\:\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{solve} \\ $$$${a}\int\frac{\mathrm{ln}\:{x}}{{x}−{b}}{dx} \\ $$$$\mathrm{and}\:\mathrm{then}\:\mathrm{insert} \\ $$$$\mathrm{I}\:\mathrm{get}\:{a}\int\frac{\mathrm{ln}\:{x}}{{x}−{b}}{dx}={a}\left(\mathrm{ln}\:\left(−{b}\right)\:\mathrm{ln}\:\mid{x}+{b}\mid\:−\mathrm{Li}_{\mathrm{2}} \:\left(\frac{{x}}{{b}}+\mathrm{1}\right)\right)+{C} \\ $$$$\mathrm{but}\:\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{sure}\:\mathrm{if}\:\mathrm{this}\:\mathrm{is}\:\mathrm{right}\:\mathrm{for}\:{a},\:{b}\:\notin\mathbb{R} \\ $$

Commented by Dwaipayan Shikari last updated on 02/Dec/20

((4x^2 )/((1+x^4 )))=(2/((x^2 −i)))+(2/(x^2 +i)) =(1/( (√i)))((1/((x−(√i))))−(1/((x+(√i)))))+(1/( i(√i)))((1/(x−i(√i)))−(1/(x+i(√i))))

$$\frac{\mathrm{4}{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{4}} \right)}=\frac{\mathrm{2}}{\left({x}^{\mathrm{2}} −{i}\right)}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} +{i}}\:=\frac{\mathrm{1}}{\:\sqrt{{i}}}\left(\frac{\mathrm{1}}{\left({x}−\sqrt{{i}}\right)}−\frac{\mathrm{1}}{\left({x}+\sqrt{{i}}\right)}\right)+\frac{\mathrm{1}}{\:{i}\sqrt{{i}}}\left(\frac{\mathrm{1}}{{x}−{i}\sqrt{{i}}}−\frac{\mathrm{1}}{{x}+{i}\sqrt{{i}}}\right) \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com