$\displaystyle \int f(x) dx$

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, $\displaystyle f(x) = 12x-9$.

The antiderivative is  $\displaystyle F(x) = 6x^2-9x + C$.

The integral is equal to

$\displaystyle 2x^2+\frac{\arctan(\frac{x}{4})}{4}+C$

and was found using the following rules:

$\displaystyle \int x^ndx=\frac{x^{n+1}}{n+1}+C$, $\displaystyle \int \frac{dx}{x^2+a^2}=\frac{1}{a}\arctan(\frac{x}{a})+C$

To integrate this expression, use u substitution:

$\displaystyle u=x^2-1; du=2xdx; \frac{1}{2}du=xdx$

Now, substitute in everything and rewrite the expression: 

$\displaystyle \frac{1}{2}\int u^{\frac{1}{2}}du$

Integrate and remember to raise the exponent by 1 and then put that result on the denominator: 

$\displaystyle (\frac{1}{2})(\frac{u^{\frac{3}{2}}}{\frac{3}{2}})$

Simplify and sub back in your initial expression:

$\displaystyle \frac{u^{\frac{3}{2}}}{3}=\frac{(x^2-1)^{\frac{3}{2}}}{3}$

Remember to add C because it is an indefinite integral:
$\displaystyle \frac{(x^2-1)^{\frac{3}{2}}}{3}+C$.

To integrate this expression, use u substitution:

$\displaystyle u=x^2-4; du=2xdx$

Now, substitute everything in and rewrite the expression:

$\displaystyle \frac{u^{\frac{3}{2}}}{\frac{3}{2}}$

Now, simplify and sub back in your initial expression:

$\displaystyle \frac{2}{3}(x^2-4)^{\frac{3}{2}}$

Don't forget to add a C because it is an indefinite integral:

$\displaystyle \frac{2}{3}(x^2-4)^{\frac{3}{2}}+C$.

First, chop up this fraction into three separate terms so you can more easily integrate:

$\displaystyle \int x^2-4x+3dx$

Now, integrate. Remember to add one to the exponent and also put that result on the denominator:

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

Simplify and add a C because it is an indefinite integral:

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

To integrate, remember to raise the exponent by 1 and then put that result on the denominator:

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

Remember to add a C because it is an indefinite integral:

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

$\displaystyle \int f(x) dx$

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, $\displaystyle f(x) = 13i$.

The antiderivative is  $\displaystyle F(x) = 13ix +C$.

$\displaystyle \int f(x) dx$

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, $\displaystyle f(x) = \frac{x}{3-x}$.

The antiderivative is  $\displaystyle F(x) = -3 \ln(3-x) - x+C$.

$\displaystyle \int f(x) dx$

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, $\displaystyle f(x) = e^{2x-5}$.

The antiderivative is  $\displaystyle F(x) = \frac{e^{2x-5} }{2} +C$.

$\displaystyle \int f(x) dx$

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, $\displaystyle f(x) = \sin (5x-3)$.

The antiderivative is  $\displaystyle F(x) = -\frac{ \cos(5x-3) }{5} +C$.