Integrating square root function 

I am calculating the length of a curve. I have:
$$x = t \sin(t), y = t \cos(t)$$
Integrating over the interval [0, 1], I get to this step, which I have verified is correct:
$$\int_0^1 \sqrt{t^2 + 1} \;\mathrm{d}t$$
The answer I get is: $\dfrac{2}{3}\cdot(t^2+1)^{\tfrac{3}{2}}$ over interval [0,1].
But my textbook has answer: [ $\dfrac{1}{2} \cdot t \cdot \sqrt{t^2+1} + \dfrac{1}{2} \ln(t + \sqrt{t^2 + 1})$ ] over interval [0,1].
I really don't understand what I am doing wrong.
Thank you for your help.
EDIT: Sorry I'm not used to MathJax, I'm trying to fix it haha!
 A: Does your textbook cover trigonometric substitution?   If so, start there (most texts these days cover it as a separate technique of integration, but it is closely related to u-substitution.  
If you follow how $t=\tan\theta$ leads to $\mathrm{d}t=\sec^2\theta \;\mathrm{d}\theta$ (it's just a derivative written a little funny), you're halfway there. 
Then   $\int\sqrt{t^2+1}\;\mathrm{d}t=\int\sqrt{(\tan\theta)^2+1}(\sec^2\theta \;\mathrm{d}\theta)=\int\sec^3\theta \;\mathrm{d}\theta$ , by trig identities.    Work out this integral first.  
For the limits, you can either reverse the substitution using $\theta=\tan^{-1} t$ to rewrite the answer back in terms of $t$ (this is back-substituting) or use the facts that $\theta=0$ when $t=0$ and $\theta=\tfrac \pi 4$ when $t=1$ to evaluate the $\int\sec^3\theta \;\mathrm{d}\theta$ integral as a definite integral with limits $0$ and $\tfrac \pi 4$ (this is converting the limits).
A: What you need here is a trigonometric substitution -- in this case, $t=\tan\theta$, so $t^2+1=\sec^2\theta$ and $\mathrm{d}t=\sec^2\theta \;\mathrm{d}\theta$. 
This leaves you with the integral $\int\sec^3\theta \;\mathrm{d}\theta$, which you can tackle using other techniques. 
You can either convert the limits to limits in terms of $\theta$ or (as seems to be indicated by the book's answer you gave) back-substitute to get an expression in terms of $t$ and plug in the given limits. 
I usually recommend converting the limits for $\theta$ because the back-substitution step is another possible source of algebra errors, but they both give the same answer.
A: One can also use hyperbolic trigonometric functions.
With $\sinh(x)=\frac 1 2(e^x-e^{-x})$ and $\cosh(x)= \frac 1 2(e^x+e^{-x})$ one has $\cosh^2(u)-\sinh^2(u)=1$ and
$$
t= \sinh(u) \Rightarrow dt = \cosh(u) du
$$
leaving
$$
 \int\limits_0^1 \sqrt{t^2 + 1} dt = 
\int\limits_0^{\sinh^{-1}(1)} \cosh^2(u)du= 
[\sinh(u)\cosh(u)]_0^{\sinh^{-1}(1)} - \int\limits_0^{\sinh^{-1}(1)} \sinh^2(u)du
=
\frac{1}{2}\left[\sinh(u)\cosh(u) +u \right]_0^{\sinh^{-1}(1)}.
$$
$\sinh^{-1}$ can be easily solved from the definition of $\sinh$, giving 
$$
\sinh^{-1}(y)=\ln(y + \sqrt{1+y^2}).
$$
A: This integral is tabulated so normally one wouldn't solve it by hand.
I'll add another non-trigonometric way of computing this; integration by parts:
$$\int\sqrt{1+t^2}dt=t\sqrt{1+t^2}-\int\frac{t^2}{\sqrt{1+t^2}}dt$$
$$\int\sqrt{1+t^2}dt=t\sqrt{1+t^2}-\int\frac{1+t^2-1}{\sqrt{1+t^2}}dt$$
$$\int\sqrt{1+t^2}dt=t\sqrt{1+t^2}-\int\sqrt{1+t^2}dt+\int\frac{1}{\sqrt{1+t^2}}dt$$
Rearrange:
$$2\int\sqrt{1+t^2}dt=t\sqrt{1+t^2}+\int\frac{1}{\sqrt{1+t^2}}dt$$
The second term on the right is something you may recognize as $\operatorname{arsinh}(t)$ (similar to the one with minus you know to be inverse sine), which is expressible with logarithms. Even if you don't know that, you extracted the first term correctly, and you may proceed to integrate the second term with another method.
