First, chop up the fraction into three simplified terms:
$\displaystyle \int x^2+4x-2dx$

Now integrate.

$\displaystyle \frac{x^3}{3}+\frac{4x^2}{2}-2x=\frac{x^3}{3}+2x^2-2x$

Now add a C because it is an indefinite integral:

$\displaystyle \frac{x^3}{3}+2x^2-2x+C$

Separate integral into two separate integrals:

$\displaystyle \int cos^{2}x+x\sqrt{x^{2}+1}dx=\int cos^{2}xdx+\int x\sqrt{x^{2}+1}dx$  

Solve $\displaystyle \int cos^{2}xdx$ first. 

Use power-reducing identity to simplify integral: $\displaystyle \int \frac{1+cos(2x)}{2}dx$.

Factor  $\displaystyle \frac{1}{2}$ out of the integral: $\displaystyle \int \frac{1}{2}+\frac{cosx}{2}dx$

Separate into two integrals: $\displaystyle \frac{1}{2}\int 1dx+\frac{1}{2}\int cos (2x)dx=\frac{1}{2}x+\frac{1}{2}\int cos (2x)dx$.

Use the following substitution for the second integral: $\displaystyle u=2x$ $\displaystyle du=2$ $\displaystyle \frac{1}{2}du=dx$

Plug in substitution and solve:  $\displaystyle \frac{1}{4}\int cosudu=\frac{1}{4}sinu+C=\frac{1}{4}sin(2x)+C$  $\displaystyle \frac{1}{4}sinu=\frac{1}{4}sin(2x)+C$.

Therefore: $\displaystyle \int cos^{2}x=\frac{1}{2}x+\frac{1}{4}sin(2x)+C$

Solve $\displaystyle \int x\sqrt{x^{2}+1}dx$

Make the following substitution: $\displaystyle u=x^{2}+1$  $\displaystyle du=2xdx$   $\displaystyle \frac{1}{2}du=dx$. Plug in substitution and solve: $\displaystyle \frac{1}{2}\int \sqrt{u}du=(\frac{2}{3})\frac{1}{2}(x^{2}+1)^{\frac{3}{2}}+C=\frac{1}{3}(x^{2}+1)^{\frac{3}{2}}+C$

Combine answers to two original integrals: $\displaystyle \int cos^{2}xdx+\int x\sqrt{x^{2}+1}dx=\frac{1}{2}x+\frac{1}{4}sin(2x)+\frac{1}{3}(x^{2}+1)^{\frac{3}{2}}+C$

$\displaystyle \int cos^{2}x+x\sqrt{x^{2}+1}dx=\frac{1}{2}x+\frac{1}{4}sin(2x)+\frac{1}{3}(x^{2}+1)^{\frac{3}{2}}+C$

$\displaystyle \int f(x) dx$

In order to evaluate this integral, first find the antiderivative of $\displaystyle f(x).$

If $\displaystyle f(x) = a$ then $\displaystyle F(x) = ax + C$

If $\displaystyle f(x) = x a$ then $\displaystyle F(x) = \frac{x^{a+1}}{a+1} + C$

If $\displaystyle f(x) = \frac{1}{x}$ then $\displaystyle F(x) = ln(x) + C$

If $\displaystyle f(x) = e ^x$ then $\displaystyle F(x) = e^x + C$

If $\displaystyle f(x) = \cos(x)$ then $\displaystyle F(x) = \sin(x) + C$

If $\displaystyle f(x) = \sin(x)$ then $\displaystyle F(x) = - \cos(x) + C$

If $\displaystyle f(x) = \sec^2 (x)$ then $\displaystyle F(x) = \tan(x) + C$

In this case, $\displaystyle f(x) = 44x^{43}$.

The antiderivative is  $\displaystyle F(x) = x^{44 }+ C$.

$\displaystyle \int f(x) dx$

In order to evaluate this integral, first find the antiderivative of $\displaystyle f(x).$

If $\displaystyle f(x) = a$ then $\displaystyle F(x) = ax + C$

If $\displaystyle f(x) = x a$ then $\displaystyle F(x) = \frac{x^{a+1}}{a+1} + C$

If $\displaystyle f(x) = \frac{1}{x}$ then $\displaystyle F(x) = ln(x) + C$

If $\displaystyle f(x) = e ^x$ then $\displaystyle F(x) = e^x + C$

If $\displaystyle f(x) = \cos(x)$ then $\displaystyle F(x) = \sin(x) + C$

If $\displaystyle f(x) = \sin(x)$ then $\displaystyle F(x) = - \cos(x) + C$

If $\displaystyle f(x) = \sec^2 (x)$ then $\displaystyle F(x) = \tan(x) + C$

In this case, $\displaystyle f(x) = \frac{1}{4x-19}$.

The antiderivative is  $\displaystyle F(x) = \frac{4x-19}{4} + C$.

$\displaystyle \int f(x) dx$

In order to evaluate this integral, first find the antiderivative of $\displaystyle f(x).$

If $\displaystyle f(x) = a$ then $\displaystyle F(x) = ax + C$

If $\displaystyle f(x) = x a$ then $\displaystyle F(x) = \frac{x^{a+1}}{a+1} + C$

If $\displaystyle f(x) = \frac{1}{x}$ then $\displaystyle F(x) = ln(x) + C$

If $\displaystyle f(x) = e ^x$ then $\displaystyle F(x) = e^x + C$

If $\displaystyle f(x) = \cos(x)$ then $\displaystyle F(x) = \sin(x) + C$

If $\displaystyle f(x) = \sin(x)$ then $\displaystyle F(x) = - \cos(x) + C$

If $\displaystyle f(x) = \sec^2 (x)$ then $\displaystyle F(x) = \tan(x) + C$

In this case, $\displaystyle f(x)= e^{8x-43}$.

The antiderivative is  $\displaystyle F(x) = \frac{e^{8x-43}}{8} + C$.

$\displaystyle \int f(x) dx$

In order to evaluate this integral, first find the antiderivative of $\displaystyle f(x).$

If $\displaystyle f(x) = a$ then $\displaystyle F(x) = ax + C$

If $\displaystyle f(x) = x a$ then $\displaystyle F(x) = \frac{x^{a+1}}{a+1} + C$

If $\displaystyle f(x) = \frac{1}{x}$ then $\displaystyle F(x) = ln(x) + C$

If $\displaystyle f(x) = e ^x$ then $\displaystyle F(x) = e^x + C$

If $\displaystyle f(x) = \cos(x)$ then $\displaystyle F(x) = \sin(x) + C$

If $\displaystyle f(x) = \sin(x)$ then $\displaystyle F(x) = - \cos(x) + C$

If $\displaystyle f(x) = \sec^2 (x)$ then $\displaystyle F(x) = \tan(x) + C$

In this case, $\displaystyle f(x)= \cos(12)$.

The antiderivative is  $\displaystyle F(x) = \cos(12)x + C$.

$\displaystyle \int f(x) dx$

In order to evaluate this integral, first find the antiderivative of $\displaystyle f(x).$

If $\displaystyle f(x) = a$ then $\displaystyle F(x) = ax + C$

If $\displaystyle f(x) = x a$ then $\displaystyle F(x) = \frac{x^{a+1}}{a+1} + C$

If $\displaystyle f(x) = \frac{1}{x}$ then $\displaystyle F(x) = ln(x) + C$

If $\displaystyle f(x) = e ^x$ then $\displaystyle F(x) = e^x + C$

If $\displaystyle f(x) = \cos(x)$ then $\displaystyle F(x) = \sin(x) + C$

If $\displaystyle f(x) = \sin(x)$ then $\displaystyle F(x) = - \cos(x) + C$

If $\displaystyle f(x) = \sec^2 (x)$ then $\displaystyle F(x) = \tan(x) + C$

In this case, $\displaystyle f(x)= \sin(10x-3)$.

The antiderivative is  $\displaystyle F(x) = -\frac{\cos(10x-3)}{10} + C$.

$\displaystyle \int_{a}^{b}f(x) \ dx$

In order to evaluate this integral, first find the antiderivative of $\displaystyle f(x).$

In this case, $\displaystyle f(x)= \cos(12)$.

The antiderivative is  $\displaystyle F(x) = \cos(12)x$.

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative: 

$\displaystyle \int_{1}^{2}\cos (12) \ dx = \left ( \cos(12)x \right )_{1}^{2}$

$\displaystyle = \cos(12)(2) - \left ( \cos(12)(1) \right )$

$\displaystyle = 0.98$

$\displaystyle \int_{a}^{b}f(x) \ dx$

In order to evaluate this integral, first find the antiderivative of $\displaystyle f(x).$

In this case, $\displaystyle f(x)= \sin(10x-3)$.

The antiderivative is  $\displaystyle F(x) = -\frac{\cos(10x-3)}{10}$.

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative: 

$\displaystyle \int_{1}^{2} \sin(10x-3) \ dx = \left ( -\frac{\cos(10x-3)}{10} \right )_{1}^{2}$

$\displaystyle = -\frac{\cos(10(2)-3)}{10} - \left ( -\frac{\cos(10-3)}{10} \right )$

$\displaystyle = 0.004$