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bình luận
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Cần giúp
giải chỗ nguyên hàm dùng vi phân là ra luôn rồi.
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sửa đổi
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nguyên hàm 12
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$\int\limits\frac{dx}{\sqrt{(x-\frac{5}{2})^2-(\frac{1}{2})^2}}=-\int\limits\frac{dx}{\sqrt{(\frac{1}{2})^2-(x-\frac{5}{2})^2}}$Đặt $x-\frac{5}{2}=\frac{1}{2}sint\Rightarrow t=arcsin(2x-5)$$\Rightarrow \left\{ \begin{array}{l} dx=\frac{1}{2}cost.dt\\ \sqrt{(\frac{1}{2})^2-(x-\frac{5}{2})^2}=\frac{1}{2} cost\end{array} \right.$$\Rightarrow -\int\limits du=-t+C=-arcsin(2x-5)+C$
$\int\limits\frac{dx}{\sqrt{(x-\frac{5}{2})^2-(\frac{1}{2})^2}}=-\int\limits\frac{dx}{\sqrt{(\frac{1}{2})^2-(x-\frac{5}{2})^2}}$Đặt $x-\frac{5}{2}=\frac{1}{2}sint\Rightarrow t=arcsin(2x-5)$$\Rightarrow \left\{ \begin{array}{l} dx=\frac{1}{2}cost.dt\\ \sqrt{(\frac{1}{2})^2-(x-\frac{5}{2})^2}=\frac{1}{2} cost\end{array} \right.$$\Rightarrow -\int\limits dt=-t+C=-arcsin(2x-5)+C$
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sửa đổi
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nguyên hàm 12
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Đợi xíu nhá
$\int\limits\frac{dx}{\sqrt{(x-\frac{5}{2})^2-(\frac{1}{2})^2}}=-\int\limits\frac{dx}{\sqrt{(\frac{1}{2})^2-(x-\frac{5}{2})^2}}$Đặt $x-\frac{5}{2}=\frac{1}{2}sint\Rightarrow t=arcsin(2x-5)$$\Rightarrow \left\{ \begin{array}{l} dx=\frac{1}{2}cost.dt\\ \sqrt{(\frac{1}{2})^2-(x-\frac{5}{2})^2}=\frac{1}{2} cost\end{array} \right.$$\Rightarrow -\int\limits du=-t+C=-arcsin(2x-5)+C$
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sửa đổi
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nguyên hàm 12
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$\int\limits\frac{dx}{\sqrt{(x-\frac{5}{2})^2-(\frac{1}{2})^2}}=ln\left| {\frac{x-3}{x-2}} \right|+C$
Đợi xíu nhá
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sửa đổi
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nguyên hàm 12
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$\int\limits\frac{dx}{\sqrt{(x-2).(x-3)}}=\int\limits\frac{(x-2)-(x-3)}{\sqrt{(x-2).(x-3)}}dx$$=\int\limits\sqrt{x-3}dx-\int\limits\sqrt{x-2}dx$$=\frac{2}{3}\sqrt{(x-3)^3}-\frac{2}{3}\sqrt{(x-2)^3}+C$
$\int\limits\frac{dx}{\sqrt{(x-\frac{5}{2})^2-(\frac{1}{2})^2}}=ln\left| {\frac{x-3}{x-2}} \right|+C$
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sửa đổi
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nguyên hàm 12
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$\int\limits\frac{dx}{\sqrt{(x-2).(x-3)}}=\int\limits\frac{(x-2)-(x-3)}{\sqrt{(x-2).(x-3)}}dx$$=\int\limits\sqrt{x-3}dx-\int\limits\sqrt{x-2}dx$$=\frac{2}{3}\sqrt{(x-3)^3}-\frac{2}{3}\sqrt{(x-2)^3}$
$\int\limits\frac{dx}{\sqrt{(x-2).(x-3)}}=\int\limits\frac{(x-2)-(x-3)}{\sqrt{(x-2).(x-3)}}dx$$=\int\limits\sqrt{x-3}dx-\int\limits\sqrt{x-2}dx$$=\frac{2}{3}\sqrt{(x-3)^3}-\frac{2}{3}\sqrt{(x-2)^3}+C$
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sửa đổi
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nguyên hàm 12
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$\int\limits\frac{dx}{\sqrt{(x-2).(x-3)}}=\int\limits\frac{(x-2)-(x-3)}{\sqrt{(x-2).(x-3)}}dx$$=\int\limits\sqrt{x-3}dx-\int\limits\sqrt{x-2}dx$$=\frac{2}{3}(\sqrt{(x-3)^3}-\frac{2}{3}(\sqrt{(x-2)^3}$
$\int\limits\frac{dx}{\sqrt{(x-2).(x-3)}}=\int\limits\frac{(x-2)-(x-3)}{\sqrt{(x-2).(x-3)}}dx$$=\int\limits\sqrt{x-3}dx-\int\limits\sqrt{x-2}dx$$=\frac{2}{3}\sqrt{(x-3)^3}-\frac{2}{3}\sqrt{(x-2)^3}$
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sửa đổi
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nguyên hàm 12
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$\int\limits\frac{dx}{\sqrt{(x-2).(x-3)}}=\int\limits\frac{(x-2)-(x-3)}{\sqrt{(x-2).(x-3)}}dx$$=\int\limits\frac{dx}{\sqrt{x-3}}-\int\limits\frac{dx}{\sqrt{x-2}}$$=2\sqrt{x-3}-2\sqrt{x-2}+C$
$\int\limits\frac{dx}{\sqrt{(x-2).(x-3)}}=\int\limits\frac{(x-2)-(x-3)}{\sqrt{(x-2).(x-3)}}dx$$=\int\limits\sqrt{x-3}dx-\int\limits\sqrt{x-2}dx$$=\frac{2}{3}(\sqrt{(x-3)^3}-\frac{2}{3}(\sqrt{(x-2)^3}$
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