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 90589 by M±th+et£s last updated on 24/Apr/20

$$\int \frac{1}{x+\sqrt{x^2+x+1}}dx$$

Commented by mathmax by abdo last updated on 24/Apr/20

$$I=\int \frac{dx}{x+\sqrt{x^2+x+1}}=\int \frac{dx}{x+\sqrt{{(x+\frac{1}{2})}^2+\frac{3}{4}}}$$$$changementx+\frac{1}{2}=\frac{\sqrt{3}}{2}sh\left(t\right)\Rightarrow$$$$I=\int \frac{1}{\frac{\sqrt{3}}{2}sh\left(t\right)-\frac{1}{2}+\frac{\sqrt{3}}{2}ch\left(t\right)}\times \frac{\sqrt{3}}{2}ch\left(t\right)dt$$$$=\sqrt{3}\int \frac{ch\left(t\right)dt}{\sqrt{3}sh\left(t\right)-1+\sqrt{3}ch\left(t\right)}=\sqrt{3}\int \frac{\frac{e^t+e^{-t}}{2}}{\sqrt{3}\frac{e^t-e^{-t}}{2}-1+\sqrt{3}\frac{e^t+e^{-t}}{2}}dt$$$$=\sqrt{3}\int \frac{e^t+e^{-t}}{\sqrt{3}(e^t-e^{-t})+\sqrt{3}(e^t+e^{-t})-2}dt$$$${=}_{e^t=u}\sqrt{3}\int \frac{u+u^{-1}}{\sqrt{3}(u-u^{-1})+\sqrt{3}(u+u^{-1})-2}\frac{du}{u}$$$$=\sqrt{3}\int \frac{u^2+1}{\sqrt{3}(u^3-u)+\sqrt{3}(u^3+u)-2u^2}du$$$$=\sqrt{3}\int \frac{u^2+1}{2\sqrt{3}{u}^3-2u^2}du=\frac{\sqrt{3}}{2}\int \frac{u^2+1}{u^2(\sqrt{3}u-1)}duletdecompose$$$$F\left(u\right)=\frac{u^2+1}{u^2(\sqrt{3}u-1)}\Rightarrow F\left(u\right)=\frac{a}{u}+\frac{b}{u^2}+\frac{c}{\sqrt{3}u-1}$$$$b=-1andc=4\Rightarrow F\left(u\right)=\frac{a}{u}-\frac{1}{u^2}+\frac{4}{\sqrt{3}u-1}$$$${lim}_{u\to +\mathrm{\infty }}uF\left(u\right)=\frac{1}{\sqrt{3}}=a+\frac{4}{\sqrt{3}}\Rightarrow a=\frac{3}{\sqrt{3}}=\sqrt{3}\Rightarrow$$$$F\left(u\right)=\frac{\sqrt{3}}{u}-\frac{1}{u^2}+\frac{4}{\sqrt{3}u-1}\Rightarrow$$$$\int F\left(u\right)du=\sqrt{3}ln\mid u\mid +\frac{1}{u}+\frac{4}{\sqrt{3}}ln\mid \sqrt{3}u-1\mid +C$$$$=\sqrt{3}ln\mid e^t\mid +e^{-t}+\frac{4}{\sqrt{3}}ln\mid \sqrt{3}{e}^t-1\mid +C$$$$t=argsh\left(\frac{2x+1}{\sqrt{3}}\right)=ln(\frac{2x+1}{\sqrt{3}}+\sqrt{1+{\left(\frac{2x+1}{\sqrt{3}}\right)}^2})\Rightarrow$$$$\int F\left(u\right)du=\sqrt{3}(\frac{2x+1}{\sqrt{3}}+\sqrt{1+{\left(\frac{2x+1}{\sqrt{3}}\right)}^2})+\frac{1}{\frac{2x+1}{\sqrt{3}}+\sqrt{1+{\left(\frac{2x+1}{\sqrt{3}}\right)}^2}}$$$$+\frac{4}{\sqrt{3}}ln\mid \sqrt{3}(\frac{2x+1}{\sqrt{3}}+\sqrt{1+{\left(\frac{2x+1}{\sqrt{3}}\right)}^2})-1\mid +C$$$$I=\frac{\sqrt{3}}{2}\int F\left(u\right)du$$$$$$

Commented by M±th+et£s last updated on 25/Apr/20

$$godblessyou$$

Commented by turbo msup by abdo last updated on 25/Apr/20

$$youarewelcomesir.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com