Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 34228 by abdo imad last updated on 03/May/18

find the value of ∫_0 ^1 (x^2 /(1+x^4 ))dx

$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{4}} }{dx} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 03/May/18

let I=∫(x^2 /(1+x^(4 ) ))dx =∫(1/(x^2 +(1/x^2 )))dx =(1/2)∫(((1+(1/x^2 ))+(1−(1/x^2 )))/(x^2 +(1/x^2 )))dx =(1/2)∫(((1+(1/x^2 )))/((x−(1/x))^2 +2))+(1/2)∫(((1−(1/x^2 )))/((x+(1/x))^2 −2)) =(1/2)∫((d(x−(1/x)))/((x−(1/x))^2 +2))+(1/2)∫((d(x+(1/x)))/((x+(1/x))^2 −2)) =(1/2).(1/(√2)).tan^(−1) {(((x−(1/x)))/(√2))}+(1/2).(1/(2(√2))).ln∣(((√2) −(x+(1/x)))/((√2)+(x+(1/x))) now given intregal has upper limit 1 and lower limit limit 0 put the limit

$${let}\:{I}=\int\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{4}\:} }{dx} \\ $$$$=\int\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)+\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}{{x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}{\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left({x}−\frac{\mathrm{1}}{{x}}\right)}{\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left({x}+\frac{\mathrm{1}}{{x}}\right)}{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}.{tan}^{−\mathrm{1}} \left\{\frac{\left({x}−\frac{\mathrm{1}}{{x}}\right)}{\sqrt{\mathrm{2}}}\right\}+\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}.{ln}\mid\frac{\sqrt{\mathrm{2}}\:−\left({x}+\frac{\mathrm{1}}{{x}}\right)}{\sqrt{\mathrm{2}}+\left({x}+\frac{\mathrm{1}}{{x}}\right.} \\ $$$${now}\:{given}\:{intregal}\:{has}\:{upper}\:{limit}\:\mathrm{1}\:{and}\:{lower}\:{limit} \\ $$$${limit}\:\mathrm{0} \\ $$$${put}\:{the}\:{limit} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com