When integrating, we do the opposite of a derivative.  You increase the exponent by one and divide the function by that new power.  Since this is an indefinite integral, we have to add a $\displaystyle +C$ at the end of the equation.

$\displaystyle \int 3x^2+2x^2+4dx=x^3+x^2+4x+C.$

The antiderivative of $\displaystyle sin(x)=-cos(x)$.  The antiderivative of $\displaystyle cos(x)=sin(x)$.  Remember, this is the opposite of a derivative.  Therefore, for our integral, we have:

$\displaystyle \int 5sin(x)-2cos(x)\;dx=-5cos(x)-2sin(x)+C$.

Using integration by parts:

$\displaystyle \int( e^{-x}*3x) dx= 3x*-e^{-x}-\int 3e^{-x}dx= -3xe^{-x}+3e^{-x}$

                              $\displaystyle =-3xe^{-x}+3e^{-x}=3e^{-x}(1-x)$

To determine the integral we just do $\displaystyle u$ substitution: 

$\displaystyle \int e^{({1+i})t} dt= \frac{1}{1+i}e^{(1+i)t}=F(t)$

By the fundamental theorem of calculus: 

$\displaystyle \int_{0}^{2\pi} e^{(1+i)t} dt = F(2\pi)-F(0)=\frac{1}{1+i}*(e^{(1+i)2\pi}-1)$

$\displaystyle re(\frac{1}{1+i}*(e^{(1+i)2\pi}-1))=\frac{1}{2}*(e^{2\pi}(cos(2\pi)-1))$

                                                  $\displaystyle =\frac{1}{2}[e^{2\pi}-1]$

Doing integration by parts twice: 

$\displaystyle \int x^2e^x dx=x^2e^x-\int2xe^x dx=x^2e^x-(2xe^x-\int 2e^x dx)$

                     $\displaystyle =x^2e^x-2xe^x+2x^x+C= e^x(x^2-2x+2) +C$  

$\displaystyle \int sin(x^2)*x dx$

Using $\displaystyle u$ substitution,

$\displaystyle u=x^2$

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} u=2x$

$\displaystyle \int sin(x^2)*x dx =\frac{1}{2}\int sin(u) du=\frac{1}{2}(-cos(u)) +C$ 

$\displaystyle \int sin(x^2)*x dx= -\frac{1}{2}cos(x^2) +C$

Upon early inspection of this problem, two things may be seen immediately: a trigonometric function and a composite function. One may notice that $\displaystyle 2x$ is the derviative of $\displaystyle x^2$, this urges us to use the u-substitution method.

Let $\displaystyle u=x^2, du=(2x) dx$, therefore the problem may be rewritten as:

$\displaystyle \int(sec^2(u)) du$, this is a known trigonometric integral to be $\displaystyle tan(u)+C$, when plugging in for u, the final answer is:$\displaystyle tan(x^2)+C$.

Using the definition above: 

$\displaystyle \int e^tsin(t) dt=\frac{1}{2i}\int e^t(e^{it})-e^t(e^{-it}) dt$

$\displaystyle =\frac{1}{2i}[\int e^{t(1+i)})dx-\int e^{t(1-i)}dt]$

This reduces to:

$\displaystyle \frac{1}{2i}[\int e^{t(1+i)})dt-\int e^{t(1-i)}dt]= \frac{1}{2i}[\frac{1}{1+i}e^{t(1+i)}-\frac{1}{1-i}e^{t(1-i)}] +C$

We can use integration by parts to solve this integral

Integration by parts states: $\displaystyle \int udv=uv-\int vdu$

Let u = $\displaystyle lnx$ and dv=$\displaystyle x^{3}dx$

Thus, our integral becomes:$\displaystyle \int x^{3}lnxdx=\frac{1}{4}x^{4}(lnx)-\int \frac{1}{4}x^{4}(\frac{1}{x})dx$

Which simplifies to: $\displaystyle \frac{1}{4}x^{4}-\frac{1}{4}\int x^{3}dx$, which equals :$\displaystyle \frac{1}{4}x^{4}lnx-\frac{1}{16}x^{4}+C$, giving us our answer$\displaystyle y=\frac{1}{4}x^{4}lnx-\frac{1}{16}x^{4}+C$