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I tried to use the definition: $$\displaystyle \Phi(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-{s^2}/{2}}\,\mathrm ds$$

So, according to this site: $$\int \:e^{-{x^2}/{2}}\mathrm dx=\frac{\sqrt{\pi }}{\sqrt{2}}\text{erf}\left(\frac{x}{\sqrt{2}}\right)+C$$

But by definition: $${\displaystyle \operatorname {erf} (x)={\frac {2}{\sqrt {\pi }}}\int _{0}^{x}e^{-s^{2}}\,\mathrm {d} s}$$

I do not know how to follow after the function $ erf (...) $

Maybe the value is only possible to get it through tables?

How to determine $x,x∈R^+$ such that $\Phi(x)=0,9505$?

Thank you very much.

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    I suggest you do not cross the streams. The integral's token variable should not appear in the integral's own boundaries. It serves as a source of much confusion. – 2017-02-16
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    Thanks for editing. You forgot to change the differential to the first equation. Look if it was okay now. – 2017-02-16
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    Would that be $\Phi,$ not $\varphi ?$ – 2017-02-16
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    Okay, I changed it. Thanks for letting me know. – 2017-02-16
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    Then numerical methods are required. In R statistical software, `qnorm(.9505)` returns 1.649721, where `qnorm` is $\Phi^{-1}.$ In what context does this arise? – 2017-02-16
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    @BruceET I do not quite understand what you're saying. Excuse me. – 2017-02-16
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    The reason we relied on printed tables of the standard normal distribution for years is that the integral in your first displayed equation cannot be evaluated in 'closed form'. Those tables were made by numerical integration. – 2017-02-16
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    @BruceET The $x$ can only be obtained by looking at the approximate probability in the table? Or is there any additional calculation that I can do with the information in the table itself? Thank you very much. – 2017-02-16
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    The table I have happens to show exactly .9505 in the body for x = 1.65, so that's the closest you can get. In some cases you might get slight improvements by interpolation, but not in this particular case. That the table has _exactly_ .9505 makes me wonder if this is an exercise in using printed tables. (You didn't give context. Where problem is from. What you're studying, etc. Makes it difficult to be of most help.) Nowadays people are using software increasingly, both for accuracy and convenience. (Bedtime here, hope that's it. Others may help if you have add'l questions.) – 2017-02-16
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    @BruceET Actually there is no context, the exercises ask to calculate $ Φ (x) = p $ using tables knowing $ x $, but nothing says of the inverse process, so the question "Maybe the value is only possible to get it through tables? ". Earlier I came calculating integrals each time I was asked to find a particular real number given a certain density function, reason why not hesitate and try to do the same, happening what you already know. I would choose the software calculation, but the problem is that I am preparing for an exam, so I want to do without it. Thank you very much for your help. – 2017-02-16

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If you use the definitions, you end with $$\displaystyle \Phi(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-{s^2}/{2}}\, ds=\frac{1}{2} \left(1+\text{erf}\left(\frac{x}{\sqrt{2}}\right)\right)$$ So, if you need to solve for $x$ equation $\Phi(x) =c$, the solution is $$x=\sqrt{2}\, \text{erf}^{-1}(2 c-1)$$

If you do not want to use tables, you could use the expansions given here

$$\text{erf}^{-1}(z)=\frac{1}{2}\sqrt{\pi}\left (z+\frac{\pi}{12}z^3+\frac{7\pi^2}{480}z^5+\frac{127\pi^3}{40320}z^7+\frac{4369\pi^4}{5806080}z^9+\frac{34807\pi^5}{182476800}z^{11}+\cdots\right)$$ The problem is that this expansion is very slowly convergent when the argument is large as in your case $(2c-1=0.901)$; the table below gives some results as a function of the number of terms used $$\left( \begin{array}{cc} n & x \\ 1 & 1.129 \\ 2 & 1.369 \\ 3 & 1.476 \\ 4 & 1.535 \\ 5 & 1.571 \\ 6 & 1.595 \\ 7 & 1.610 \\ 8 & 1.621 \\ 9 & 1.629 \\ 10 & 1.634 \\ 11 & 1.638 \\ 12 & 1.641 \\ 13 & 1.643 \\ 14 & 1.645 \\ 15 & 1.646 \\ 16 & 1.647 \\ 17 & 1.647 \\ 18 & 1.648 \\ 19 & 1.648 \\ 20 & 1.649 \\ 21 & 1.649 \end{array} \right)$$

You could also use the approximation $$\operatorname{erf}(z) \approx \sqrt{1 - \exp\left(-z^2\frac{\frac{4}{\pi} + az^2}{1 + az^2}\right)}$$ given in the Wikipedia page which then reduce to a quadratic equation in $z^2$ (use $a=0.147$).

For the vale you give, this would lead to $x=1.64953$ while the exact value is $x=1.64972$.