Question

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

Answer 1

\[\int\limits (3x^2 +x+4)/(x^3+x)\]

Answer 2

anyone?

Answer 3

what i dont understand is when we use Bx+c and when we use just BX

Answer 4

solution would probably make explaining a lot easier

Answer 5

http://www3.wolframalpha.com/Calculate/MSP/MSP282119g0f704g4if65h10000185c8ih78h2420gd?MSPStoreType=image/gif&s=3&w=436&h=40

Answer 6

dhatra ,I know the answer but i dont know how to solve the partial fraction part

Answer 7

http://www.intmath.com/methods-integration/11-integration-partial-fractions.php
that should give you an idea when to use what

Answer 8

I used Bx + c and got

4/x^2 + 1 / x^2+1 - x/x^2 + 1

Answer 9

sorry its 4/x

Answer 10

integrating

= 4 lnx + arctan x - (1/2) ln (x^2 + 1) + C

Answer 11

working back the partial fractions work out

Answer 12

r u there rav?

Answer 13

Rav-Doing your partial fraction decomp.
You want to use just a constant for linear factors such as:
\[\frac{3}{(x+2)}\]. Since x+2 is a linear factor.
If you have something in the form:
\[\frac{4}{(x+2)^2}\].
You want something in the form:
\[\frac{A}{(x+2)}+\frac{B}{(x+2)^2}\].
You only want to use Cx+D (since I already used A and B)
when you have an irreducible quadratic term (something that doesn't factor PERIOD)
Such as:
\[\frac{7}{x^2+2x+19}\].
You want something in the form:
\[\frac{Cx+D}{x^2+2x+19}\].
Then if you have powers of that you want to include those as well. Let me make up and example that displays all of them and show you how to proceed.
Ex.
\[\frac{7x+2}{(x-3)^3(x^2+4x+97)^2 (x^2+3x+2) (2x+1)}\].
You would factor the x^2+3x+2. Then set up a decomp that looks like this:
\[\frac{A}{(x-3)}+\frac{B}{(x-3)^2}+\frac{C}{(x-3)^3}+\frac{Dx+E}{(x^2+4x+97)}\]
\[+\frac{Fx+G}{x^2+4x+97}+\frac{H}{(x+2)}+\frac{I}{(x+1)}+\frac{J}{(2x+1)}\].
Tell me if you need more help Rav :P