How do you integrate $\int \frac{e^{\frac{1}{x^{2}}}}{x^{3}} d x$ $int e^(1/x^2)/x^3dx$ ?

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How do you integrate $\int \frac{e^{\frac{1}{x^{2}}}}{x^{3}} d x$ $int e^(1/x^2)/x^3dx$ ?

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Answer 1 · 2017-01-07T21:24:47

$- \frac{e^{1 / x^{2}}}{2} + C$ $-e^(1//x^2)/2+C$

Explanation:

$\int \frac{e^{1 / x^{2}}}{x^{3}} d x$ $inte^(1//x^2)/x^3dx$

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

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

$= \int e^{1 / x^{2}} (\frac{1}{x^{3}} d x) = - \frac{1}{2} \int e^{1 / x^{2}} (- \frac{2}{x^{3}} d x) = - \frac{1}{2} \int e^{u} d u$ $=inte^(1//x^2)(1/x^3dx)=-1/2inte^(1//x^2)(-2/x^3dx)=-1/2inte^udu$

The integral of $e^{u}$ $e^u$ is itself:

$= - \frac{1}{2} e^{u} = - \frac{e^{1 / x^{2}}}{2} + C$ $=-1/2e^u=-e^(1//x^2)/2+C$

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How do you integrate $\int \frac{e^{\frac{1}{x^{2}}}}{x^{3}} d x$ $int e^(1/x^2)/x^3dx$ ?

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