integral of Trigonometric function $\begin{align}
(1)\quad &\int\dfrac{1}{\sqrt{x^4+1}}\mathrm dx\\
(2)\quad &\int\dfrac{\sin x}{x}\mathrm dx
\end{align}$
How can I solve these types of integrals?
 A: These integrals are not "elementary". The first one is basically an elliptic integral of the first kind, which can be explicitly written in terms of hypergeometric functions after some manipulations (see Euler's integral representation of hypergeometric functions).
The second one is the sine integral, which also has no representation in terms of elementary functions.
Hope this helps!
A: Bruno already mentioned that both of the OP's integrands are not expressible in terms of "usual" functions; I'll address the "how can I evaluate" portion of the question for the first integral.
Letting $x=\tan\frac{u}{2}$, $\mathrm dx=\frac{\mathrm du}{1+\cos\,u}$ (the Weierstrass substitution), we get
$$\int \frac1{(1+\cos\,u)\sqrt{1+\tan^4\frac{u}{2}}}\mathrm du$$
$$\int \frac1{\sqrt{(1+\cos\,u)^2\left(1+\left(\frac{1-\cos\,u}{1+\cos\,u}\right)^2\right)}}\mathrm du$$
which simplifies to
$$\int\frac{\mathrm du}{\sqrt{4-2\,\sin^2 u}}$$
Removing a factor of $\frac12$ from the integrand gives
$$\frac12\int\frac{\mathrm du}{\sqrt{1-\frac12\sin^2 u}}$$
which is easily recognized as the incomplete elliptic integral of the first kind:
$$\frac12 F\left(u\mid\frac12\right)$$
Undoing the initial Weierstrass substitution yield the final result,
$$\frac12 F\left(2\arctan\,x\mid\frac12\right)$$
Alternatively, one could have instead used $x=\cot\frac{u}{2}$ as the substitution, but I'll leave that as an exercise...
