Solve: $y''=(1+(y')^2)^{3/2}$ / $y(1)=\frac 15, y'(1)=\frac 34$. My attempt:
$y''=(1+(y')^2)^{3/2}$
$\displaystyle\implies \frac {y''}{(1+(y')^2)^{3/2}}=1$
$\displaystyle\implies \int_{}^{} \frac {y''}{(1+(y')^2)^{3/2}} dx=x+c$
$\displaystyle\implies \int\limits_{}^{}\frac{1}{(1+u^2)\sqrt{1+u^2}}du=x+c,$ where $  u=y'$
$\displaystyle\implies \int\limits_{}^{}\frac{\sqrt{1+u^2}}{(1+u^2)^2}du=x+c$
I am stuck here, is my approach correct and can solve the problem? Thanks for your help.

You have found that
$\displaystyle\int\frac{du}{(1+u^2)^{\frac{3}{2}}}=x+c$
Let $u=\tan\theta$ , therefore $du=\sec^2\theta d\theta$
$\displaystyle\int\frac{\sec^2\theta  d\theta}{\sec^3\theta}=x+c$
$\displaystyle\int\cos\theta=x+c$
$\sin\theta=x+c$
$\displaystyle \frac{u}{\sqrt{1+u^2}}=x+c$
where $u=y'$
We can find $c$ by the fact the $u(1)=\frac{3}{4}$
Solving we get
$\displaystyle \frac{u}{\sqrt{1+u^2}}=x-\frac{2}{5}$
Now $u=\frac{dy}{dx}$ and then our integral becomes
$\displaystyle\int dy =\int \frac{x-\frac{2}{5}}{\sqrt{1-(x-\frac{2}{5}})^2}dx$
which can be easily solved using substitution.
The final answer would be
$(y-1)^2+(x-\frac{2}{5})^2=1$
