How do you integrate $int e^(1/x^2)/x^3dx$ ?

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How do you integrate $int e^(1/x^2)/x^3dx$ ?

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Answer 1 · 2017-01-07T21:24:47

$-e^(1//x^2)/2+C$

Explanation:

$inte^(1//x^2)/x^3dx$

Let $u=1/x^2$ . Differentiating this, we see that $du=-2/x^3dx$ .

We currently have $1/x^3dx$ in the integrand, so we need to multiply by $-2$ in the integral and $-1/2$ to balance this on the exterior of the integral.

$=inte^(1//x^2)(1/x^3dx)=-1/2inte^(1//x^2)(-2/x^3dx)=-1/2inte^udu$

The integral of $e^u$ is itself:

$=-1/2e^u=-e^(1//x^2)/2+C$

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How do you integrate $int e^(1/x^2)/x^3dx$ ?

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How do you integrate int e^(1/x^2)/x^3dxint e^(1/x^2)/x^3dx?

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How do you integrate $int e^(1/x^2)/x^3dx$ ?