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ບອກຄວາມແຕກຕ່າງ w.r.t. x

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5 ບັນຫາທີ່ຄ້າຍຄືກັນກັບ:

ບັນຫາທີ່ຄ້າຍຄືກັນຈາກWeb Search

https://www.quora.com/Can-the-integral-displaystyle-int-sqrt-frac-x-1-x-2-mathrm-d-x-be-calculated

https://www.quora.com/How-do-we-integrate-int-frac-7-sqrt-x-x-3-dx

*A2A \displaystyle\begin{equation}\begin{split}\int \dfrac{7^{\sqrt{x}}}{x^3}\mathrm{dx}&=2\int \dfrac{7^{\sqrt{x}}}{2\sqrt{x}\cdot x^{\frac{5}{2}}}\mathrm{dx}\\\text{Let }y=\sqrt{x}&\implies \mathrm{dy}=\dfrac{1}{2\sqrt{x}}\mathrm{dx}\\&=2\int \dfrac{7^y}{y^5}\mathrm{dy}\\\text{Let }u=7^y&\implies \mathrm{du}=7^y\ln 7\mathrm{dy}\\\text{And }\mathrm{dv}=y^{-5}\mathrm{dy}&\implies v=-\dfrac{y^{-4}}{4}\\&=2\left[-\dfrac{7^y y^{-4}}{4}+\int \dfrac{7^yy^{-4}\ln 7}{4}\mathrm{dy}\right]\\\text{Let }u=7^y\ln 7&\implies \mathrm{du}=7^y(\ln 7)^2\mathrm{dy}\\\text{And }\mathrm{dv}=\dfrac{y^{-4}}{4}\mathrm{dy}&\implies v=-\dfrac{y^{-3}}{12}\\&=2\left[-\dfrac{7^yy^{-4}}{4}-\dfrac{7^yy^{-3}\ln 7}{12}+\int \dfrac{7^yy^{-3}(\ln 7)^2}{12}\mathrm{dy}\right]\\\text{Let }u=7^y(\ln 7)^2&\implies \mathrm{du}=7^y(\ln 7)^3\mathrm{dy}\\\text{And }\mathrm{dv}=\dfrac{y^{-3}}{12}\mathrm{dy}&\implies v=-\dfrac{y^{-2}}{24}\\&=2\left[-\dfrac{7^yy^{-4}}{4}-\dfrac{7^yy^{-3}\ln 7}{12}-\dfrac{7^yy^{-2}(\ln 7)^2}{24}+\int \dfrac{7^yy^{-2}(\ln 7)^3}{24}\mathrm{dy}\right]\\\text{Let }u=7^y(\ln 7)^3&\implies \mathrm{du}=7^y(\ln 7)^4\mathrm{dy}\\\text{And }\mathrm{dv}=\dfrac{y^{-2}}{24}&\implies v=-\dfrac{y^{-1}}{24}\\&=2\left[-\dfrac{7^yy^{-4}}{4}-\dfrac{7^yy^{-3}\ln 7}{12}-\dfrac{7^yy^{-2}(\ln 7)^2}{24}-\dfrac{7^yy^{-1}(\ln 7)^3}{24}+\int \dfrac{7^yy^{-1}(\ln 7)^4}{24}\mathrm{dy}\right]\\&=-\dfrac{7^y}{2y^4}-\dfrac{7^y\ln 7}{6y^3}-\dfrac{7^y(\ln 7)^2}{12y^2}-\dfrac{7^y(\ln 7)^3}{12y}+\dfrac{(\ln 7)^4}{12}\int\dfrac{7^y}{y}\mathrm{dy}\\&=-\dfrac{7^y}{2y^4}-\dfrac{7^y\ln 7}{6y^3}-\dfrac{7^y(\ln 7)^2}{12y^2}-\dfrac{7^y(\ln 7)^3}{12y}+\dfrac{(\ln 7)^4}{12}\int\dfrac{e^{\ln 7^y}}{y}\mathrm{dy}\\&=-\dfrac{7^y}{2y^4}-\dfrac{7^y\ln 7}{6y^3}-\dfrac{7^y(\ln 7)^2}{12y^2}-\dfrac{7^y(\ln 7)^3}{12y}+\dfrac{(\ln 7)^4}{12}\int\dfrac{e^{y\ln 7}}{y}\mathrm{dy}\\&=-\dfrac{7^y}{2y^4}-\dfrac{7^y\ln 7}{6y^3}-\dfrac{7^y(\ln 7)^2}{12y^2}-\dfrac{7^y(\ln 7)^3}{12y}+\dfrac{(\ln 7)^4}{12}\text{Ei}(y\ln 7)+C\\&=\boxed{-\dfrac{7^{\sqrt{x}}}{2x^2}-\dfrac{7^{\sqrt{x}}\ln 7}{6x\sqrt{x}}-\dfrac{7^{\sqrt{x}}(\ln 7)^2}{12x}-\dfrac{7^{\sqrt{x}}(\ln 7)^3}{12\sqrt{x}}+\dfrac{(\ln 7)^4}{12}\text{Ei}(\sqrt{x}\ln 7)+C}\end{split}\end{equation}\tag*{}

https://www.quora.com/How-do-you-integrate-displaystyle-int-frac-dx-sqrt-4x-x-2

https://socratic.org/questions/how-do-you-integrate-int-sqrt-1-x-2-xdx-using-trigonometric-substitution-1

https://math.stackexchange.com/questions/1660864/integrating-int-frac1-sqrt1-x2-dx-using-u-substitution-with-an-x

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