Calculus problem HELP!ZZ?

Question

find the integral of (x+4)/(2x^2 + 3x +2)

help!

come on no one knows how to do?

it is 2X^2

Answer 1

Use integration by parts. Let u = x + 4 and dv = 1/(2x^2 + 3x + 2) dx. Using the formula, int(u dv) = uv - int(v du), you'll need to find du and v. The derivative of u is du = 1 dx. To find the integral of dv, you'll need to manipulate the expression...
int[1/(2x^2 + 3x + 2) dx] = (1/2)int[1/(x^2 +(3/2)x + 1) dx]
= (1/2)int[1/{(x^2 + (3/2)x +(9/16)) + (1 - (9/16)} dx]
= (1/2)int[1/{(x + (3/4))^2 + (7/16)} dx]
= (1/2)*[1/(sqrt(7)/16)*arctan{(x + (3/4))/(sqrt(7)/16)}
= 8sqrt(7)/7*arctan{[16sqrt(7)(x + 3/4)]/7}

Now, you can plug and chug into the formula.
uv = (x + 4)*8sqrt(7)/7*arctan{[16sqrt(7)(x + 3/4)]/7} and
int(v du) = int[8sqrt(7)/7*arctan{[16sqrt(7)(x + 3/4)]/7}*1 dx]
= 8sqrt(7)/7*int[arctan{[16sqrt(7)(x + 3/4)]/7} dx]
= 8sqrt(7)/7*16sqrt(7)(x + 3/4)/7*arctan{[16sqrt(7)(x + 3/4)]/7} - (1/2)ln|1 + {[16sqrt(7)(x + 3/4)]/7}^2|
= 128(x + 3/4)/7*arctan{[16sqrt(7)(x + 3/4)]/7} - (1/2)ln|1 + {[16sqrt(7)(x + 3/4)]/7}^2|

Now finally,
uv - int(v du) = (x + 4)*8sqrt(7)/7*arctan{[16sqrt(7)(x + 3/4)]/7} - 128(x + 3/4)/7*arctan{[16sqrt(7)(x + 3/4)]/7} + (1/2)ln|1 + {[16sqrt(7)(x + 3/4)]/7}^2| + C
That should do it for you. Very tedious, but doable with some substitutions and tables. Good luck.

Answer 2

Lobosito has an interesting lead, but is the denominator 2x^2 + 3x + 2 or 2(x^2 + 3x + 2)?

If 2x^2 + 3x + 2, then it has thus far resisted any of my attempts to use a substitution, complete the square on the denominator, or fractionally decompose it into a sum of two linear rational functions.

I think this one is a "refer to the table of integration forms" problem.

edit:
Lee is correct. I didn't even think of integration by parts. d'oh!

Answer 3

int (x+4)/2(x^2+3x+2)=

1/2 int (x+4)/(x+1)(x+2)

now use that 1/(x+1)(x+2)=1/(x+1)-1/(x+2)