# How can I calculate the integral?

How can I calculate the following integral??

$$\int \frac{1}{t}u'(t) dt$$

I thought that I could it as followed:

$$\int \frac{1}{t}u'(t) dt=\frac{1}{t}u(t)+\int \frac{1}{t^2}u(t) dt$$

but I don't know how to continue...

Or is there an other way to calculate it??

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AFAIK, this has no closed form. – Hakim Jun 15 '14 at 18:29

## 1 Answer

This integral has no closed form since we don't know what this function $u'(t)$ can be. For example, take $u(t)=-\cos t$, then $u'(t)=\sin t$, and thus we will get: $$\int\dfrac{\sin t}{t}\mathrm dt\,,$$ which is known to not have an antiderivative expressible in terms of elementary functions. So if a closed form for your integral existed then that would mean that a closed form for this since integral exists, which isn't possible.

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Ahaa...Ok! I am solving an hyperbolic system of PDE using the method of characteristics and I wanted to write the following expression as a total derivative. $$\left ( \frac{\partial{u_1}}{t}-\frac{x}{t}\frac{\partial{u_2}}{\partial{t}} \right )+\left (( 1-xt)\frac{\partial{u_1}}{\partial{x}}+\left (x^2-\frac{x}{t}\right )\frac{\partial{u_2}}{\partial{x}} \right )=0$$ So, can I not write this as a total derivative?? – Mary Star Jun 15 '14 at 19:00
I mean, when we had: $$\displaystyle{(\frac{\partial{u_1}}{\partial{t}}+\sqrt{ab} \frac{\partial{u_1}}{\partial{x}})+\sqrt{\frac{a}{b}}(\frac{ \partial{u_2}}{ \partial{t} }+\sqrt{ab} \frac{\partial{u_2}}{\partial{x}})=0}$$ we could write it as: $$\displaystyle{\frac{\partial}{\partial{t}}(u_1+\sqrt{\frac{a}{b}}u_2)+\sqrt{ab‌​} \frac{\partial}{\partial{x}}(u_1+\sqrt{\frac{a}{b}}u_2)=0}$$ which is the derivative: $$\frac{d}{d{t}}(u_1+\sqrt{\frac{a}{b}}u_2)$$ when $\displaystyle{\frac{dx}{dt}\sqrt{ab}}$ – Mary Star Jun 15 '14 at 19:05
@MaryStar I'm sorry I can't help since I don't have any knowledge about PDEs. Maybe you can consider writing a new question. – Hakim Jun 15 '14 at 19:08
Ok! I've written it as a new question... – Mary Star Jun 16 '14 at 15:28
@MaryStar I wasn't addressing you of course. ;-) – Hakim Jun 16 '14 at 15:31