3 replies to this topic

[tritanngo99]

Tìm nguyên hàm: $I=\int (1+x+\frac{1}{x})e^{x-\frac{1}{x}}dx$

[stuart clark]

$$I = \int \left(1+x+\frac{1}{x}\right)e^{x-\frac{1}{x}}dx = \int \left(1+\frac{1}{x}+\frac{1}{x^2}\right)e^{x-\frac{1}{x}+\ln(x)}dx$$

 

$x-\frac{1}{x}+\ln(x) = t\;,$ Th $\displaystyle \left(1+\frac{1}{x}+\frac{1}{x^2}\right)dx = dt$

 

$$ I = \int e^tdt = e^t+\mathcal{C} = e^{x-\frac{1}{x}+\ln(x)}+\mathcal{C} = xe^{\left(x-\frac{1}{x}\right)}+\mathcal{C}$$

[tritanngo99]

$$I = \int \left(1+x+\frac{1}{x}\right)e^{x-\frac{1}{x}}dx = \int \left(1+\frac{1}{x}+\frac{1}{x^2}\right)e^{x-\frac{1}{x}+\ln(x)}dx$$

 

$x-\frac{1}{x}+\ln(x) = t\;,$ Th $\displaystyle \left(1+\frac{1}{x}+\frac{1}{x^2}\right)dx = dt$

 

$$ I = \int e^tdt = e^t+\mathcal{C} = e^{x-\frac{1}{x}+\ln(x)}+\mathcal{C} = xe^{\left(x-\frac{1}{x}\right)}+\mathcal{C}$$

Another way:

$I=\int (1+x+\frac{1}{x})e^{x-\frac{1}{x}}dx=\int e^{x-\frac{1}{x}}dx+\int (x+\frac{1}{x})e^{x-\frac{1}{x}}dx(1)$

Consider $\int (x+\frac{1}{x})e^{x-\frac{1}{x}}dx=\int x(1+\frac{1}{x^2})e^{x-\frac{1}{x}}dx$.

$u=x;dv=(1+\frac{1}{x^2})e^{x-\frac{1}{x}}\implies du=dx \text{ and choose } v=e^{x-\frac{1}{x}}$

$\implies \int (x+\frac{1}{x})e^{x-\frac{1}{x}}dx=xe^{x-\frac{1}{x}}-\int e^{x-\frac{1}{x}}dx(2)$.

From $(1)\text{ and } (2)\implies I=xe^{x-\frac{1}{x}}+C$.

Edited by tritanngo99, 17-07-2016 - 18:57.

[stuart clark]

Given $$\displaystyle \int \left(1+x+\frac{1}{x}\right)\cdot e^{x-\frac{1}{x}}dx\;,$$ Now put $$x=e^{i\phi} = \cos \phi+i\sin \phi.$$

 

Then $$\displaystyle x^{-1}=e^{-i\phi} = \cos \phi-i\sin \phi.$$ So we get $$\displaystyle \left(x+\frac{1}{x}\right)=2\cos \phi.$$ 

 

and $$\displaystyle \left(x-\frac{1}{x}\right)=2i\sin \phi.$$ and $dx = ie^{i\phi}d\phi.$

 

So Integral convert into $$\displaystyle \int \left(1+2\cos \phi\right)\cdot e^{2i\sin \phi} \cdot ie^{i\phi}d\phi = \int i(1+2\cos \phi)\cdot e^{i(\phi+2\sin \phi)}d\phi.$$

 

Now Let $$\displaystyle i(\phi+2\sin \phi)=t\;,$$ Then $i(1+2\cos \phi)d\phi = dt.$

 

So we get $$\displaystyle \int e^{t}dt = e^{t}+\mathcal{C} = e^{i(\phi+2\sin \phi)}+\mathcal{C} = e^{i\phi}\cdot e^{2i\sin \phi}+\mathcal{C} = x\cdot e^{x-\frac{1}{x}}+\mathcal{C}$$.

 

So $$\displaystyle \int \left(1+x+\frac{1}{x}\right)\cdot e^{x-\frac{1}{x}}dx = x\cdot e^{x-\frac{1}{x}}+\mathcal{C}.$$