Definite integral $\int_0^1\sqrt{x^2+1}\, dx$. Use trig substitution? I have an integral: 

$$\int_0^1\sqrt{x^2+1}\, dx$$

but I have gotten stuck. Here's the work I have done already:

I'm not sure where to go from here. Using a trig identity doesn't seem like it would work. Integration by parts doesn't work (at least I think so)
Any ideas on what I should do next?
Thank you!

Hint:) 
Since $\cosh^2u=\sinh^2u+1$ so let $x=\sinh u$.

It's better to do this
$$\int\sec^3t dt=\int\dfrac{\cos t}{(1-\sin^2t)^2}dt$$

Edit:
\begin{align}
I  
&= \int\sec^3t dt \\
&= \int\dfrac{1}{\cos^3t} dt \\
&= \int\dfrac{\cos t}{\cos^4t} dt \\
&= \int\dfrac{\cos t}{(1-\sin^2t)^2}dt
\end{align}
with substitution $\sin t=u$:
$$I=\int\dfrac{du}{(1-u^2)^2}$$
with fraction decomposition
\begin{align}
\dfrac{1}{(1-u^2)^2} 
&= \dfrac{A}{1-u}+\dfrac{B}{1+u}+\dfrac{C}{(1-u)^2}+\dfrac{D}{(1+u)^2} \\
&= \dfrac{(A+B+C+D)+(A-B+2C-2D)u+(-A-B+C+D)u^2+(-A+B)u^3}{(1-u^2)^2}
\end{align}
equivalency of coefficients in numerator shows
\begin{cases}
A+B+C+D=1,\\
A-B+2C-2D=0,\\
-A-B+C+D=0,\\
-A+B=0.
\end{cases}
so $A=B=C=D=\dfrac14$ and thus with $u=\sin t$
\begin{align}
I  
&= \dfrac14\int\left(\dfrac{1}{1-u}+\dfrac{1}{1+u}+\dfrac{1}{(1-u)^2}+\dfrac{1}{(1+u)^2}\right)du \\
&= \dfrac14\left(-\ln(1-u)+\ln(1+u)+\dfrac{1}{1-u}-\dfrac{1}{1+u}\right)+C \\
&= \dfrac14\left(\ln\dfrac{1+u}{1-u}+\dfrac{2u}{1-u^2}\right)+C \\
&= \dfrac14\left(\ln\dfrac{1+\sin t}{1-\sin t}+\dfrac{2\sin t}{1-\sin^2t}\right)+C \\
\end{align}
substitution $\sin^2t=\dfrac{x^2}{1+x^2}$ gives us
$$\color{blue}{\int\sqrt{x^2+1}\,dx=\dfrac12\left(\ln(x+\sqrt{x^2+1})+x\sqrt{x^2+1}\right)+C}$$
