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  \(\displaystyle x^2\).

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

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

\(\displaystyle \sin(x)\) is a special function.

The indefinite integral is \(\displaystyle \int \sin(x) {\mathrm{d} x}=-\cos(x)+c\).

Even though it is a special function, we still need 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  \(\displaystyle x^2\).

To solve this problem, we can use the anti-power rule or reverse power rule. We raise the exponent on the variables by one and divide by the new exponent.

For this problem, we'll treat \(\displaystyle 8\) as \(\displaystyle 8x^0\) since anything to the zero power is one.

\(\displaystyle \int 6x^2+8{\mathrm{d} x}=\frac{6x^{2+1}}{2+1}+\frac{8x^{0+1}}{0+1}+c\)

Since the derivative of any constant is \(\displaystyle 0\), when we take the indefinite integral, we add a \(\displaystyle c\) to compensate for any constant that might be there.

From here we can simplify.

\(\displaystyle \int 6x^2+8{\mathrm{d} x}=\frac{6x^{3}}{3}+\frac{8x^{1}}{1}+c\)

\(\displaystyle \int 6x^2+8{\mathrm{d} x}=2x^3+8x+c\)