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Question Number 53477 by maxmathsup by imad last updated on 22/Jan/19

$$letf\left(a\right)={\int }_0^1\frac{dt}{\sqrt{x+a}+3}$$$$1)calculatef\left(a\right)$$$$2)findalso{\int }_0^1\frac{dt}{\sqrt{x+a}{(\sqrt{x+a}+3)}^2}$$$$3)findthevaluesofintegrals{\int }_0^1\frac{dt}{\sqrt{x+1}+3}and{\int }_0^1\frac{dt}{\sqrt{x+1}{(\sqrt{x+1}+3)}^2}$$

Commented by Abdo msup. last updated on 23/Jan/19

$$1)wehavef\left(a\right)={\left[2(\sqrt{x+a}+3)\right]}_0^1$$$$=2(\sqrt{a+1}-\sqrt{a})$$$$2)wehave{f}^{{}^{\prime }}\left(a\right)={\int }_0^1\frac{\partial }{\partial a}\left(\frac{1}{\sqrt{x+a}+3}\right)dx$$$$={\int }_0^1-\frac{{(\sqrt{x+a}+3)}^{,}}{{(\sqrt{x+a}+3)}^2}dx=-{\int }_0^1\frac{1}{2\left(\sqrt{x+a}\right){(\sqrt{x+a}+3)}^2}dx$$$$\Rightarrow {\int }_0^1\frac{dx}{\left(\sqrt{x+a}\right){(\sqrt{x+a}+3)}^2}=-2f^{{}^{\prime }}\left(a\right)$$$$=-4{(\sqrt{a+1}-\sqrt{a})}^{{}^{\prime }}=-4(\frac{1}{2\sqrt{a+1}}-\frac{1}{2\sqrt{a}})$$$$=2(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{a+1}})=2\frac{\sqrt{a+1}-\sqrt{a}}{\sqrt{a^2+a}}$$$$3)wehave{\int }_0^1\frac{dt}{\sqrt{x+1}+3}=f\left(1\right)=2(\sqrt{2}-1)$$$${\int }_0^1\frac{dx}{\left(\sqrt{x+1}\right){(\sqrt{x+1}+3)}^2}=-2f^{{}^{\prime }}\left(1\right)=2\frac{\sqrt{2}-1}{\sqrt{2}}$$$$=\sqrt{2}(\sqrt{2}-1)=2-\sqrt{2}.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jan/19

$$1)\frac{1}{\sqrt{x+a}+3}\times \mid t{\mid }_0^1=\frac{1}{\sqrt{x+a}+3}$$$$sirithinkdtismistyped...itshouldbedx...$$

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