I'm totally lost on the following, need to find the indefinite integral...
$\displaystyle
\int\frac{3x^{4}-5}{x^{2}}dx
$
$\displaystyle
\int\frac{4e^{x}}{5+e^{x}}dx
$
$\displaystyle
\int\frac{x+3}{x^{2}+6x+8}dx
$
I'm totally lost on the following, need to find the indefinite integral...
$\displaystyle
\int\frac{3x^{4}-5}{x^{2}}dx
$
$\displaystyle
\int\frac{4e^{x}}{5+e^{x}}dx
$
$\displaystyle
\int\frac{x+3}{x^{2}+6x+8}dx
$
No there was no other info given by my teacher other than these these problems, and some of the basic rules used. I just can't figure out how to utilize those rules for these particular problems. The list of rules include the general power rule, etc...
Write this as,$\displaystyle
\int\frac{3x^{4}-5}{x^{2}}dx
$
$\displaystyle \int \frac{3x^4}{x^2} - \frac{5}{x^2} dx$
Use the substitution $\displaystyle t=5+e^x$.$\displaystyle
\int\frac{4e^{x}}{5+e^{x}}dx
$
Use the substitution $\displaystyle t=x^2+6x+8$.$\displaystyle
\int\frac{x+3}{x^{2}+6x+8}dx
$
I thought you said you had been given the power rule?
As TPH suggested:
[tex]\int\frac{3x^{4}-5}{x^{2}}dx = \int \left ( 3x^2 - \frac{5}{x^2}\right ) dx <-- This is just simple division
$\displaystyle = \int (3x^2 - 5x^{-2}) dx$
Now for the power rule:
$\displaystyle = 3 \frac{1}{3}x^3 - 5 \frac{1}{-1}x^{-1} + C$
$\displaystyle = x^3 + 5x^{-1} + C$
$\displaystyle = x^3 + \frac{5}{x} + C$
-Dan
Again, as TPH suggested use $\displaystyle t = 5 + e^x$ which implies
$\displaystyle dt = e^x dx$
$\displaystyle dx = e^{-x}dt = \frac{1}{t - 5} dt$
So
$\displaystyle \int \frac{4e^{x}}{5+e^{x}}dx = \int \frac{4 \cdot (t - 5)}{t} \cdot \frac{1}{t - 5} dt = \int \frac{4}{t}~dt$
Now you take it from here.
-Dan