Question

integral 1 to 2 of 1/x cubed + x squared dx

Answer 1

Is it:
\[\large \int\limits\frac{1}{x^{3}} +x^{2} dx\]
?

Answer 2

x cubed + x squared are all under the dividing line

Answer 3

Is it:
\[\large \int\limits\frac{1}{x^{3}+x^{2}} \]
?

Answer 4

yes, all of that dx

Answer 5

Partial fractions would work..

Answer 6

partial fractions are the way forward:

\[\int\limits{\frac{1}{x^3 +x^2}}dx = \int\limits{\frac{1}{x^2(x+1)}dx} = \int\limits{\frac{A}{x}} + \frac{B}{x^2} + \frac{C}{x+1}dx\]

Answer 7

so:

\[\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1} = \frac{1}{x^2(x+1)}\]

\[Ax(x+1) + B(x+1) + Cx^2 = 1\]

let x = 0
in that equation to find B

let x = -1
to find C

let x = 1 and sub in values of B and C to find A

Answer 8

yeah i always seem to end up missing something at some point when converting to P.F

Answer 9

I usually never guess PF.

Answer 10

@Nastech are you ok with that?

Answer 11

yo @Nastech

Answer 12

i think you can also do this by writing it as

x^-3 + x^2 and integrating term by term

Answer 13

oh - sorry - i misread the question

Answer 14

:) that would be the case if it was (1/x^3) + x^2

Answer 15

yes

Answer 16

yes I've got the answer thanks to you all for your assistance

Answer 17

awesome