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Question Number 85162 by mathmax by abdo last updated on 19/Mar/20

$$1)find\int ln(\sqrt{x}+\sqrt{x+1})dx$$$$2)calculate{\int }_0^{1}ln(\sqrt{x}+\sqrt{x+1})dx$$

Commented by mathmax by abdo last updated on 01/Apr/20

$$1)letf\left(t\right)=\int ln(t+\sqrt{x}+\sqrt{x+1})dx$$$$wehave{f}^{{}^{\prime }}\left(t\right)\int \frac{1}{t+\sqrt{x}+\sqrt{x+1}}dx{=}_{\sqrt{x}=u}\int \frac{2udu}{t+u+\sqrt{u^2+1}}$$$${=}_{u=sh\left(\alpha \right)}2\int \frac{sh\left(\alpha \right)ch\left(\alpha \right)d\alpha }{t+sh\left(\alpha \right)+ch\left(\alpha \right)}=\int \frac{sh\left(2\alpha \right)}{t+sh\alpha +ch\alpha }d\alpha$$$$=\int \frac{\frac{e^{2\alpha }-e^{-\alpha }}{2}}{t+e^{\alpha }}d\alpha {=}_{e^{\alpha }=z}\frac{1}{2}\int \frac{z^2-z^{-2}}{t+z}\frac{dz}{z}$$$$=\frac{1}{2}\int \frac{z^2-z^{-2}}{z(t+z)}dz=\frac{1}{2}\int \frac{z^4-1}{z^3(t+z)}dzletdecompose$$$$F\left(z\right)=\frac{z^4-1}{z^3(z+t)}\Rightarrow F\left(z\right)=\frac{z^4-1}{z^4+{tz}^3}=\frac{z^4+{tz}^3-{tz}^3-1}{z^4+{tz}^3}=1-\frac{{tz}^3+1}{z^4+{tz}^3}$$$$w\left(z\right)=\frac{{tz}^3+1}{z^3(z+t)}=\frac{a}{z}+\frac{b}{z^2}+\frac{c}{z^3}+\frac{d}{z+t}$$$$c=\frac{1}{t},d=\frac{1-t^4}{-t^3}=\frac{t^4-1}{t^3}\Rightarrow w\left(z\right)=\frac{a}{z}+\frac{b}{z^2}+\frac{1}{{tz}^3}+\frac{t^4-1}{t^3(z+t)}$$$${lim}_{z\to +\mathrm{\infty }}zw\left(z\right)=t=a+d\Rightarrow a=t-d=t-\frac{t^4-1}{t^3}=\frac{1}{t^3}\Rightarrow$$$$w\left(z\right)=\frac{1}{t^{3}z}+\frac{b}{z^2}+\frac{1}{{tz}^3}+\frac{t^4-1}{t^3(z+t)}$$$$w\left(1\right)=1=\frac{1}{t^3}+b+\frac{1}{t}+\frac{t^4-1}{t+1}=\frac{1}{t^3}+\frac{1}{t}+\frac{t^4-1}{t+1}$$$$=\frac{t+t^3}{t^4}+\frac{t^4-1}{t+1}=\frac{(t+t^3)(t+1)+t^8-t^4}{t^4(t+1)}$$$$=\frac{t^2+t+t^4+t^3+t^8-t^4}{t^4(t+1)}=\frac{t^8+t^3+t^2+t}{t^4(t+1)}=\frac{t^7+t^2+t+1}{t^3(t+1)}\Rightarrow$$$$b=1-\frac{t^7+t^2+t+1}{t^3(t+1)}wehaveF\left(z\right)=1-\frac{a}{z}-\frac{b}{z^2}-\frac{c}{z^3}-\frac{d}{z+t}\Rightarrow$$$$\int F\left(z\right)dz=z-aln\mid z\mid +\frac{b}{z}+\frac{c}{2z^2}-dln\mid z+t\mid +C...becontinued...$$

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