I have been trying to solve this integral $\int\frac{(x-1)e^{1/x}dx}{x}$ I used WolframAlpha to solve it but it doesn't show the process. The solution is $e^{1/x}{x} + constant$
What is the integral of $\int\frac{(x-1)e^{1/x}dx}{x}$?
3 Answers
We can integrate this by separating the integrand and integrating by parts: $\begin{align} \int\frac{(x-1)e^{1/x}}{x}dx &=\int e^{1/x}dx-\int \frac{1}{x} e^{1/x}dx\\ &=\int e^{1/x}dx+\int x\left(\frac{d}{dx}e^{1/x}\right)dx\\ &= \int e^{1/x}dx + xe^{1/x}- \int e^{1/x}dx\\ &= xe^{1/x} + C \end{align}$
$x=\frac1u,dx=-\frac1{u^2}du$
$-\int\dfrac{(\frac1u-1)\frac{e^u}{u^2}du}{\frac1u}$
Bring the minus inside and multiply top and bottom by $u$. Bring the remaining $\frac1u$ inside the parentheses.
$\int(\frac1u-\frac1{u^2})e^udu=$
$\int\frac1ud(e^u)+e^ud(\frac1u)=\int d(\frac{e^u}u)=\frac{e^u}u+C=xe^{\frac1x}+C$
Rewrite the integral:
$\int\frac{(1-x)e^{1/x}}x\,dx=\int\left(1-\frac1x\right)e^{1/x}\,dx=\int e^{1/x}\,dx-\int\,\frac1xe^{1/x}\,dx\;.$
Now try computing the first integral by parts, with $u=e^{1/x}$ and $dv=dx$. You get $du=-\frac1{x^2}e^{1/x}dx$ and $v=x$, yielding
$\int e^{1/x}\,dx=xe^{1/x}+\int\frac1xe^{1/x}\,dx\;.$
But it follows immediately from this that
$\int e^{1/x}\,dx-\int\,\frac1xe^{1/x}\,dx=xe^{1/x}$
up to the usual constant of integration.
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0Just kidding, Prof. Could I make you some questions in chat about the thing I'm doing on the floor function? – 2012-08-19