How do you integrate $\int \frac{e^{\frac{3}{x}}}{x^{2}} d x$ $int e^(3/x)/x^2dx$ from $[0, 1]$ $[0,1]$ ?

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How do you integrate $\int \frac{e^{\frac{3}{x}}}{x^{2}} d x$ $int e^(3/x)/x^2dx$ from $[0, 1]$ $[0,1]$ ?

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Answer 1 · 2017-01-14T23:02:10

The integral does not converge.

Explanation:

Notice that $\frac{e^{\frac{3}{x}}}{x^{2}}$ $e^(3/x)/x^2$ is undefined at $x = 0$ $x=0$ , so we are working with an improper integral. Instead of having a bound of $0$ $0$ , we will take the limit of the integral as some variable approaches $0$ $0$ .

Before doing so, first find the general antiderivative of the function.

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

Let $u = \frac{3}{x}$ $u=3/x$ so $d u = - \frac{3}{x^{2}} d x$ $du=-3/x^2dx$ .

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

So:

$int_0^1e^(3/x)/x^2dx=lim_(brarr0^+)int_b^1e^(3/x)/x^2dx=lim_(brarr0^+)[-1/3e^(3/x)]_b^1$

Evaluating:

$=-1/3e^3+lim_(brarr0^+)1/3e^(3/b)$

As $brarr0$ from the right, we see that $3/brarroo$ . Thus $lim_(brarr0^+)1/3e^(3/b)=oo$ as well and the integral will not converge.

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How do you integrate $\int \frac{e^{\frac{3}{x}}}{x^{2}} d x$ $int e^(3/x)/x^2dx$ from $[0, 1]$ $[0,1]$ ?

1 Answer

Explanation:

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