The integral of \(\displaystyle \frac{1}{x}\) is \(\displaystyle \ln \left | x \right |+C\).  The constant 3 is simply multiplied by the integral.  

To integrate \(\displaystyle \cos(x)\sin(x)\), we need to get the two equations in terms of each other. We are going to use "u-substitution" to create a new variable, \(\displaystyle u\), which will equal \(\displaystyle \cos(x)\).

Now, if \(\displaystyle u=\cos(x)\), then 

\(\displaystyle \frac{\mathrm{d}u }{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}\cos(x)=-sin(x)\) 

Multiply both sides by \(\displaystyle \mathrm d {x}\) to get the more familiar: 

\(\displaystyle {\mathrm{d} u}=-\sin(x)\mathrm{d}x\)

Note that our \(\displaystyle \mathrm d{u}=-\sin(x)\), and our original equation was asking for a positive \(\displaystyle \sin(x)\).

That means if we want \(\displaystyle \int\cos(x)\sin(x)\) in terms of \(\displaystyle u\), it looks like this:

\(\displaystyle \int u\cdot (-\mathrm d{u}})\)

Bring the negative sign to the outside:

\(\displaystyle -\int u\mathrm d{u}\).

We can use the power rule to find the integral of \(\displaystyle u\):

\(\displaystyle -(\frac{1}{2}u^2+c)\)

Since we said that \(\displaystyle u=\cos(x)\), we can plug that back into the equation to get our answer:

\(\displaystyle -(\frac{1}{2}\cos^2(x)+c)\)

In this case we have a rational function as \(\displaystyle \frac{N(x)}{D(x)}\), where

\(\displaystyle N(x)=1\)

and

 \(\displaystyle D(x)=\frac{1}{x^2-16}\)

\(\displaystyle D(x)\) can be written as a product of linear factors:

\(\displaystyle \frac{1}{x^2-16}=\frac{1}{(x-4)(x+4)}\equiv \frac{A}{x-4}+\frac{B}{x+4}\)

It is assumed that A and B are certain constants to be evaluated. Denominators can be cleared by multiplying both sides by (x - 4)(x + 4). So we get:

\(\displaystyle 1=A(x+4)+B(x-4)\)

First we substitute x = -4 into the produced equation:

\(\displaystyle 1=-8B\Rightarrow B=-\frac{1}{8}\)

Then we substitute x = 4 into the equation:

\(\displaystyle 1=8A\Rightarrow A=\frac{1}{8}\)

Thus:

\(\displaystyle \frac{1}{(x-4)(x+4)}=\frac{1}{8}\frac{1}{x-4}-\frac{1}{8}\frac{1}{x+4}\)

Hence:

\(\displaystyle \int \frac{dx}{x^2-16}=\int(\frac{1}{8}\frac{1}{x-4}-\frac{1}{8}\frac{1}{x+4})dx\)

\(\displaystyle =\frac{1}{8}ln\left | x-4 \right |-\frac{1}{8}ln\left |x+4 \right |+c\)

\(\displaystyle =\frac{1}{8}(ln\left | x-4 \right |-ln\left |x+4 \right |)+c\)

\(\displaystyle =\frac{1}{8}ln\left | \frac{x-4}{x+4}\right |+c\)

To solve for the indefinite integral, we can use the reverse power rule. We raise the power of the exponents by one and divide by that new exponent. For this problem, that would look like:

\(\displaystyle \int x^2+5x{\mathrm{d} x}=\frac{x^{2+1}}{2+1}+\frac{5x^{1+1}}{1+1}+c\)

Remember, when taking an integral, definite or indefinite, we always add \(\displaystyle +c\), as there could be a constant involved.

\(\displaystyle \int x^2+5x{\mathrm{d} x}=\frac{x^{3}}{3}+\frac{5x^{2}}{2}+c\)

\(\displaystyle \int x^2+5x{\mathrm{d} x}=\frac{1}{3}x^3+\frac{5}{2}x^2+c\)

To solve for the indefinite integral, we can use the reverse power rule. We raise the power of the exponents by one and divide by that new exponent. For this problem, that would look like:

\(\displaystyle \int 5x+8{\mathrm{d} x}=\frac{5x^{1+1}}{1+1}+\frac{8x^{0+1}}{0+1}+c\)

Remember, when taking an integral, definite or indefinite, we always add \(\displaystyle +c\), as there could be a constant involved.

\(\displaystyle \int 5x+8{\mathrm{d} x}=\frac{5x^{2}}{2}+\frac{8x^{1}}{1}+c\)

\(\displaystyle \int 5x+8{\mathrm{d} x}=\frac{5}{2}x^2+8x +c\)

To solve for the indefinite integral, we can use the reverse power rule. We raise the power of the exponents by one and divide by that new exponent.

We're going to treat \(\displaystyle 5\) as \(\displaystyle 5x^0\), as anything to the zero power is one.

For this problem, that would look like:

\(\displaystyle \int x^3+2x+5{\mathrm{d} x}=\frac{x^{3+1}}{3+1}+\frac{2x^{1+1}}{1+1}+\frac{5x^{0+1}}{0+1}+c\)

Remember, when taking an integral, definite or indefinite, we always add \(\displaystyle +c\), as there could be a constant involved.

\(\displaystyle \int x^3+2x+5{\mathrm{d} x}=\frac{x^{4}}{4}+\frac{2x^{2}}{2}+\frac{5x^{1}}{1}+c\)

\(\displaystyle \int x^3+2x+5{\mathrm{d} x}=\frac{1}{4}x^4+x^2+5x+c\)

To find the indefinite integral of our expression, we can use the reverse power rule.

To use the reverse power rule, we raise the exponent of the \(\displaystyle x\) by one and then divide by that new exponent.

First we need to realize that \(\displaystyle 2x=2x^1\). From there we can solve:

\(\displaystyle \int2x{\mathrm{d} x}=\frac{2x^{1+1}}{1+1}+c\)

When taking an integral, be sure to include a \(\displaystyle +c\) at the end of everything. \(\displaystyle c\) stands for "constant". Since taking the derivative of a constant whole number will always equal \(\displaystyle 0\), we include the \(\displaystyle +c\) to anticipate the possiblity of the equation actually being \(\displaystyle x^2+5\) or \(\displaystyle x^2-100\) instead of just  \(\displaystyle x^2\).

\(\displaystyle \int2x{\mathrm{d} x}=\frac{2x^{2}}{2}+c\)

\(\displaystyle \int2x{\mathrm{d} x}=x^2+c\)

To find the indefinite integral of our equation, we can use the reverse power rule.

To use the reverse power rule, we raise the exponent of the \(\displaystyle x\) by one and then divide by that new exponent.

Remember that, when taking the integral, we treat constants as that number times \(\displaystyle x^{0}\) since anything to the zero power is \(\displaystyle 1\). For example, treat \(\displaystyle 2\) as \(\displaystyle 2x^0\).

\(\displaystyle \int x^3+2{\mathrm{d} x}=\frac{x^{3+1}}{3+1}+\frac{2x^{0+1}}{0+1}+c\)

When taking an integral, be sure to include a \(\displaystyle +c\) at the end of everything. \(\displaystyle c\) stands for "constant". Since taking the derivative of a constant whole number will always equal \(\displaystyle 0\), we include the \(\displaystyle +c\) to anticipate the possiblity of the equation actually being \(\displaystyle x^2+5\) or \(\displaystyle x^2-100\) instead of just  \(\displaystyle x^2\).

\(\displaystyle \int x^3+2{\mathrm{d} x}=\frac{x^{4}}{4}+\frac{2x^{1}}{1}+c\)

\(\displaystyle \int x^3+2{\mathrm{d} x}=\frac{1}{4}x^4+2x+c\)

To find the indefinite integral of our equation, we can use the reverse power rule.

To use the reverse power rule, we raise the exponent of the \(\displaystyle x\) by one and then divide by that new exponent.

\(\displaystyle \int \frac{1}{3}x^2{\mathrm{d} x}=\frac{1}{3}*\frac{x^{2+1}}{2+1}+c\)

When taking an integral, be sure to include a \(\displaystyle +c\). \(\displaystyle c\) stands for "constant". Since taking the derivative of a constant whole number will always equal \(\displaystyle 0\), we include the \(\displaystyle +c\) to anticipate the possiblity of the equation actually being \(\displaystyle x^2+5\) or \(\displaystyle x^2-100\) instead of just  \(\displaystyle x^2\).

\(\displaystyle \int \frac{1}{3}x^2{\mathrm{d} x}=\frac{1}{3}*\frac{x^{3}}{3}+c\)

\(\displaystyle \int \frac{1}{3}x^2{\mathrm{d} x}=\frac{1}{9}x^3+c\)

To find the indefinite integral of our equation, we can use the reverse power rule.

To use the reverse power rule, we raise the exponent of the \(\displaystyle x\) by one and then divide by that new exponent.

Remember that, when taking the integral, we treat constants as that number times \(\displaystyle x^{0}\), since anything to the zero power is \(\displaystyle 1\). Treat \(\displaystyle 18\) as \(\displaystyle 18x^0\).

\(\displaystyle \int x+18{\mathrm{d} x}=\frac{x^{1+1}}{1+1}+\frac{18x^{0+1}}{0+1}+c\)

When taking an integral, be sure to include a \(\displaystyle +c\). \(\displaystyle c\) stands for "constant". Since taking the derivative of a constant whole number will always equal \(\displaystyle 0\), we include the \(\displaystyle +c\) to anticipate the possiblity of the equation actually being \(\displaystyle x^2+5\) or \(\displaystyle x^2-100\) instead of just  \(\displaystyle x^2\).

\(\displaystyle \int x+18{\mathrm{d} x}=\frac{x^{2}}{2}+\frac{18x^{1}}{1}+c\)

\(\displaystyle \int x+18{\mathrm{d} x}=\frac{1}{2}x^2+18x+c\)