evalute: $\displaystyle \int \frac{x + 3}{x^2 + 6 x}dx$
Or you can use partial fractions:
$\displaystyle x^2 + 6x = x(x + 6)$
So let's find an A and B such that
$\displaystyle \frac{x + 3}{x^2 + 6x} = \frac{A}{x} + \frac{B}{x + 6}$
Well...
$\displaystyle \frac{A}{x} + \frac{B}{x + 6} = \frac{A(x + 6) + Bx}{x^2 + 6x} = \frac{(A + B)x + 6A}{x^2 + 6x}$
Comparing the two expressions gives me
A + B = 1
6A = 3
Giving me that $\displaystyle A = B = \frac{1}{2}$.
So
$\displaystyle \int dx \frac{x + 3}{x^2 + 6x} = \int dx \frac{1/2}{x} + \int dx \frac{1/2}{x + 6}$
Can you finish this?
-Dan
heres what i've done:
$\displaystyle \int \frac{x + 3}{x^2 + 6 x}dx =$
$\displaystyle \int \frac{x}{x^2 + 6 x}dx + \int \frac{3}{x^2 + 6 x}dx$
$\displaystyle u = x^2+6x, du = 2x+6$
$\displaystyle ln(x+6)+ \int \frac{3}{x^2 + 6 x}dx$
is this right sofar?
That's nice! never thought of doing partial fractions. substitution was the way i immediately saw to do it. and since it's one of the easiest integration techniques, i didn't bother looking for another
$\displaystyle \int \frac {x + 3}{x^2 + 6x}dx$
Let $\displaystyle u = x^2 + 6x$
$\displaystyle \Rightarrow du = 2x + 6dx = 2(x + 3) dx$
$\displaystyle \Rightarrow \frac {1}{2}du = (x + 3)dx$
So our integral becomes:
$\displaystyle \frac {1}{2} \int \frac {1}{u}du$