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 39135 by maxmathsup by imad last updated on 02/Jul/18

calculate A(λ) = ∫_0 ^λ    ((ln(x+(√(1+x^2 ))))/(√(1+x^2 ))) dx  2) calculate ∫_0 ^1  ((ln(x+(√(1+x^2 ))))/(√(1+x^2 )))dx

$${calculate}\:{A}\left(\lambda\right)\:=\:\int_{\mathrm{0}} ^{\lambda} \:\:\:\frac{{ln}\left({x}+\sqrt{\left.\mathrm{1}+{x}^{\mathrm{2}} \right)}\right.}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:{dx} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left({x}+\sqrt{\left.\mathrm{1}+{x}^{\mathrm{2}} \right)}\right.}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx} \\ $$

Commented by math khazana by abdo last updated on 03/Jul/18

let integrate by parts u^′ =(1/(√(1+x^2 ))) and v=ln(x+(√(1+x^2 )))  A(λ) =[ln^2 (x+(√(1+x^2 )))]_0 ^λ  −∫_0 ^λ     ((ln(x+(√(1+x^2 ))))/(√(1+x^2 )))dx  ⇒2A(λ)=ln^2 (λ+(√(1+λ^2 ))) ⇒  A(λ)=(1/2)ln^2 (λ +(√(1+λ^2 )))  2) ∫_0 ^1    ((ln(1+(√(1+x^2 )))/(√(1+x^2 ))))dx=A(1)=(1/2)ln^2 (2+(√2)).

$${let}\:{integrate}\:{by}\:{parts}\:{u}^{'} =\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:{and}\:{v}={ln}\left({x}+\sqrt{\left.\mathrm{1}+{x}^{\mathrm{2}} \right)}\right. \\ $$$${A}\left(\lambda\right)\:=\left[{ln}^{\mathrm{2}} \left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\right]_{\mathrm{0}} ^{\lambda} \:−\int_{\mathrm{0}} ^{\lambda} \:\:\:\:\frac{{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx} \\ $$$$\Rightarrow\mathrm{2}{A}\left(\lambda\right)={ln}^{\mathrm{2}} \left(\lambda+\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\right)\:\Rightarrow \\ $$$${A}\left(\lambda\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\lambda\:+\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\right) \\ $$$$\left.\mathrm{2}\left.\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{ln}\left(\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right.}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right){dx}={A}\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\mathrm{2}+\sqrt{\mathrm{2}}\right). \\ $$

Commented by math khazana by abdo last updated on 03/Jul/18

A(1)=(1/2)ln^2 (1+(√2))

$${A}\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Jul/18

    1)t=ln(x+(√(1+x^2 )) )    (dt/dx)=((1+((2x)/(2(√(1+x^2 )) )))/(x+(√(1+x^2 )) ))=(1/((√(1+x^2 )) ))  ∫_0 ^(ln(λ+(√(1+λ^2 )) )) tdt  ∣ (t^2 /2)∣_0 ^(ln(λ+(√(1+λ^2  ))))   (1/2){ln(λ+(√(1+λ^2 ))) }^2   2)(1/2){ln∣1+(√2) ∣}^2

$$\left.\:\:\:\:\mathrm{1}\right){t}={ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\right)\:\: \\ $$$$\frac{{dt}}{{dx}}=\frac{\mathrm{1}+\frac{\mathrm{2}{x}}{\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:}}{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:}=\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:} \\ $$$$\int_{\mathrm{0}} ^{{ln}\left(\lambda+\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\:\right)} {tdt} \\ $$$$\mid\:\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\mid_{\mathrm{0}} ^{{ln}\left(\lambda+\sqrt{\left.\mathrm{1}+\lambda^{\mathrm{2}} \:\right)}\right.} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left(\lambda+\sqrt{\left.\mathrm{1}+\lambda^{\mathrm{2}} \right)}\:\right\}^{\mathrm{2}} \right. \\ $$$$\left.\mathrm{2}\right)\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\mid\mathrm{1}+\sqrt{\mathrm{2}}\:\mid\right\}^{\mathrm{2}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com