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Question Number 74223 by mathmax by abdo last updated on 20/Nov/19

$$calculatef\left(a\right)={\int }_0^1\sqrt{x^2+ax+1}dxandg\left(a\right)={\int }_0^1\frac{xdx}{\sqrt{x^2+ax+1}}$$ $$with\mid a\mid <2$$ $$2)findthevalueof{\int }_0^1\sqrt{x^2+\sqrt{2}x+1}dxand{\int }_0^1\frac{xdx}{\sqrt{x^2+\sqrt{2}x+1}}$$

Answered by mind is power last updated on 20/Nov/19

$$f\left(a\right)={\int }_0^1\sqrt{{(x+\frac{a}{2})}^2+\frac{4-a^2}{4}}dx$$ $$let(x+\frac{a}{2})=\sqrt{\frac{4-a^2}{4}}sh\left(u\right)$$ $$\Rightarrow dx=\sqrt{\frac{4-a^2}{4}}.ch\left(u\right)du$$ $${\int }_{argsh\left(\frac{a}{\sqrt{4-a^2}}\right)}^{argsh\left(\sqrt{\frac{a+2}{2-a}}\right)}.\frac{4-a^2}{4}.{ch}^2\left(u\right)du$$ $$=\int \frac{4-a^2}{4}.\left(\frac{ch\left(2u\right)+1}{2}\right)du$$ $$=\frac{4-a^2}{4}{[\frac{sh\left(2u\right)}{4}+\frac{u}{2}]}_{argsh\left(\frac{a}{\sqrt{4-a^2}}\right)}^{argsh\left(\sqrt{\frac{a+2}{2-a}}\right)}$$ $$sh\left(2u\right)=2sh\left(u\right)ch\left(u\right)$$ $$sh\left(2argsh\left(t\right)\right)=2t.\sqrt{1+t^2}$$ $$=\frac{4-a^2}{4}[\frac{\sqrt{\frac{a+2}{2-a}}\sqrt{\frac{4}{2-a}}-\frac{a}{\sqrt{4-a^2}}\sqrt{\frac{4-a^2+a}{4-a^2}}}{2}+\frac{1}{2}argsh\left(\sqrt{\frac{a+2}{2-a}}\right)-\frac{1}{2}argsh\left(\frac{a}{\sqrt{4-a^2}}\right)=f\left(a\right)$$ $$g\left(a\right)=2f^{\prime }\left(a\right)$$ $$2)=f\left(\sqrt{2}\right),g\left(\sqrt{2}\right)$$ $$$$ $$$$ $$$$

Commented bymathmax by abdo last updated on 20/Nov/19

$$thankyousir.$$

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