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Question Number 83675 by niroj last updated on 05/Mar/20
evaluate: 2 ∫_0 ^( 2) ((√(x+1))/(x^2 +4))dx
$$ \\ $$$$\: \\ $$$$\:\:\mathrm{evaluate}: \\ $$$$\:\mathrm{2}\:\int_{\mathrm{0}} ^{\:\mathrm{2}} \:\frac{\sqrt{\mathrm{x}+\mathrm{1}}}{\mathrm{x}^{\mathrm{2}} +\mathrm{4}}\mathrm{dx} \\ $$$$\:\:\:\:\: \\ $$$$\:\: \\ $$
Commented by jagoll last updated on 05/Mar/20
(√(x+1)) = u ⇒ x^2 = (u^2 −1)^2 2x dx = 4u (u^2 −1)du dx = ((2u(u^2 −1)du)/(u^2 −1)) = 2u du x^2 +4 = (u^2 −1)^2 +4 2 ∫_1 ^(√3) ((u (2u du))/(4+(u^2 −1)^2 )) let u = sec t ⇒ du = sec t tan t dt
$$\sqrt{{x}+\mathrm{1}}\:=\:{u}\:\Rightarrow\:{x}^{\mathrm{2}} =\:\left({u}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}{x}\:{dx}\:=\:\mathrm{4}{u}\:\left({u}^{\mathrm{2}} −\mathrm{1}\right){du} \\ $$$${dx}\:=\:\frac{\mathrm{2}{u}\left({u}^{\mathrm{2}} −\mathrm{1}\right){du}}{{u}^{\mathrm{2}} −\mathrm{1}}\:=\:\mathrm{2}{u}\:{du} \\ $$$${x}^{\mathrm{2}} \:+\mathrm{4}\:=\:\left({u}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4} \\ $$$$\mathrm{2}\:\overset{\sqrt{\mathrm{3}}} {\int}_{\mathrm{1}} \:\frac{{u}\:\left(\mathrm{2}{u}\:{du}\right)}{\mathrm{4}+\left({u}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\: \\ $$$${let}\:{u}\:=\:\mathrm{sec}\:{t}\:\Rightarrow\:\mathrm{du}\:=\:\mathrm{sec}\:{t}\:\mathrm{tan}\:{t}\:{dt} \\ $$$$ \\ $$
Commented by niroj last updated on 05/Mar/20
thanks mr.jagoll try to solve if it is better tobe complete.
$${thanks}\:{mr}.{jagoll}\:{try}\:{to}\:{solve}\:{if}\:{it}\:{is}\:{better}\:{tobe}\:{complete}. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by TANMAY PANACEA last updated on 05/Mar/20
I=∫((x+1)/( (√(x+1)) ×(x^2 +4)))dx t^2 =x+1 ∫((t^2 ×2tdt)/(t(t^4 −2t^2 +1+4))) 2∫(t^2 /(t^4 −2t^2 +5))dt 2∫(dt/(t^2 +(5/t^2 )−2)) t^2 +(5/t^2 )=(t+((√5)/t))^2 −2(√5) (d/dt)(t+((√5)/t))=1−((√5)/t^2 ) t^2 +(5/t^2 )=(t−((√5)/t))^2 +2(√5) (d/dt)(t−((√5)/t))=1+((√5)/t^2 ) look (1+((√5)/t^2 ))+(1−((√5)/t^2 ))=2 now∫((1−((√5)/t^2 )+1+((√5)/t^2 ))/(t^2 +(5/t^2 )−2))dt ∫((d(t+((√5)/t)))/((t+((√5)/t))^2 −2(√5) −2))+∫((d(t−((√5)/t)))/((t−((√5)/t))^2 +2(√5) −2)) now pls use formula of ∫(dx/(x^2 −a^2 )) and ∫(dx/(x^2 +a^2 )) 2(√5) +2=((√(2(√5) +2)) )^2 ←as if a^2 2(√5) −2=((√(2(√5) −2)) )^2 so (1/(2((√(2(√5) +2)) )))ln((((t+((√5)/t))−((√(2(√5) +2)) ))/((t+((√5)/t))+((√(2(√5) +2)) ))))+(1/( (√(2(√5) −2))))tan^(−1) (((t−((√5)/t))/( (√(2(√5) −2))))) now pls replace t by (√(x+1)) and put upper and lower limit
$${I}=\int\frac{{x}+\mathrm{1}}{\:\sqrt{{x}+\mathrm{1}}\:×\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{dx} \\ $$$${t}^{\mathrm{2}} ={x}+\mathrm{1} \\ $$$$\int\frac{{t}^{\mathrm{2}} ×\mathrm{2}{tdt}}{{t}\left({t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}+\mathrm{4}\right)} \\ $$$$\mathrm{2}\int\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} +\mathrm{5}}{dt} \\ $$$$\mathrm{2}\int\frac{{dt}}{{t}^{\mathrm{2}} +\frac{\mathrm{5}}{{t}^{\mathrm{2}} }−\mathrm{2}} \\ $$$${t}^{\mathrm{2}} +\frac{\mathrm{5}}{{t}^{\mathrm{2}} }=\left({t}+\frac{\sqrt{\mathrm{5}}}{{t}}\right)^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{5}}\: \\ $$$$\frac{{d}}{{dt}}\left({t}+\frac{\sqrt{\mathrm{5}}}{{t}}\right)=\mathrm{1}−\frac{\sqrt{\mathrm{5}}}{{t}^{\mathrm{2}} } \\ $$$${t}^{\mathrm{2}} +\frac{\mathrm{5}}{{t}^{\mathrm{2}} }=\left({t}−\frac{\sqrt{\mathrm{5}}}{{t}}\right)^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{5}}\: \\ $$$$\frac{{d}}{{dt}}\left({t}−\frac{\sqrt{\mathrm{5}}}{{t}}\right)=\mathrm{1}+\frac{\sqrt{\mathrm{5}}}{{t}^{\mathrm{2}} } \\ $$$${look}\:\:\left(\mathrm{1}+\frac{\sqrt{\mathrm{5}}}{{t}^{\mathrm{2}} }\right)+\left(\mathrm{1}−\frac{\sqrt{\mathrm{5}}}{{t}^{\mathrm{2}} }\right)=\mathrm{2} \\ $$$${now}\int\frac{\mathrm{1}−\frac{\sqrt{\mathrm{5}}}{{t}^{\mathrm{2}} }+\mathrm{1}+\frac{\sqrt{\mathrm{5}}}{{t}^{\mathrm{2}} }}{{t}^{\mathrm{2}} +\frac{\mathrm{5}}{{t}^{\mathrm{2}} }−\mathrm{2}}{dt} \\ $$$$\int\frac{{d}\left({t}+\frac{\sqrt{\mathrm{5}}}{{t}}\right)}{\left({t}+\frac{\sqrt{\mathrm{5}}}{{t}}\right)^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{5}}\:−\mathrm{2}}+\int\frac{{d}\left({t}−\frac{\sqrt{\mathrm{5}}}{{t}}\right)}{\left({t}−\frac{\sqrt{\mathrm{5}}}{{t}}\right)^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{5}}\:−\mathrm{2}} \\ $$$$\boldsymbol{{now}}\:\boldsymbol{{pls}}\:\boldsymbol{{use}}\:\boldsymbol{{formula}}\:\boldsymbol{{of}}\: \\ $$$$\int\frac{{dx}}{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }\:\:{and}\:\int\frac{{dx}}{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} } \\ $$$$\mathrm{2}\sqrt{\mathrm{5}}\:+\mathrm{2}=\left(\sqrt{\mathrm{2}\sqrt{\mathrm{5}}\:+\mathrm{2}}\:\right)^{\mathrm{2}} \:\:\leftarrow{as}\:{if}\:{a}^{\mathrm{2}} \\ $$$$\mathrm{2}\sqrt{\mathrm{5}}\:−\mathrm{2}=\left(\sqrt{\mathrm{2}\sqrt{\mathrm{5}}\:−\mathrm{2}}\:\right)^{\mathrm{2}} \\ $$$${so} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\left(\sqrt{\mathrm{2}\sqrt{\mathrm{5}}\:+\mathrm{2}}\:\right)}{ln}\left(\frac{\left({t}+\frac{\sqrt{\mathrm{5}}}{{t}}\right)−\left(\sqrt{\mathrm{2}\sqrt{\mathrm{5}}\:+\mathrm{2}}\:\right)}{\left({t}+\frac{\sqrt{\mathrm{5}}}{{t}}\right)+\left(\sqrt{\mathrm{2}\sqrt{\mathrm{5}}\:+\mathrm{2}}\:\right)}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{5}}\:−\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\frac{{t}−\frac{\sqrt{\mathrm{5}}}{{t}}}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{5}}\:−\mathrm{2}}}\right) \\ $$$$\boldsymbol{{now}}\:\boldsymbol{{pls}}\:\boldsymbol{{replace}}\:\boldsymbol{{t}}\:\boldsymbol{{by}}\:\sqrt{\boldsymbol{{x}}+\mathrm{1}}\:\boldsymbol{{and}}\:\boldsymbol{{put}}\:\boldsymbol{{upper}}\: \\ $$$$\boldsymbol{{and}}\:\boldsymbol{{lower}}\:\boldsymbol{{limit}} \\ $$$$ \\ $$
Commented by niroj last updated on 05/Mar/20
you are absute amazing.nice solution..
$${you}\:{are}\:{absute}\:{amazing}.{nice}\:{solution}.. \\ $$
Commented by TANMAY PANACEA last updated on 05/Mar/20
thank you sir
$${thank}\:{you}\:{sir} \\ $$

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