Question

[(2x^3 + 3x^2 +28) / ((x+2)(x^2))]
integrate 4rm 1 to 2

Answer 1

please help me?

Answer 2

with steps if you dont mind.

Answer 3

not as bad as it seems. divide,
\[\frac{2x^3+3x^2+28}{x^3+2x^2}\] is what you need

Answer 4

ca you divide it with the steps. shown.

Answer 5

i will try but it is hard to write division here. let me write it first with paper

Answer 6

ah ok.

Answer 7

well actually you get a quotient of 2, and a remainder of
\[-x^2+28\]

Answer 8

2
_______________
x^3 + 2x^2 | 2x^3+3x^2 + 28

2x^3 + 4x^2

-x^2 + 28

Answer 9

that is the best way i can write it here

Answer 10

just like long division, you say x^3 goes in to 2x^3 twice, put the 2 up top, multiply out and subtract

Answer 11

that means v multiply because it says integrate from 1 to 2 is it?

Answer 12

meaning
\[\frac{2x^3+3x^2+28}{x^3+2x^2}=2+\frac{-x^2+28}{x^3+2x^2}\]

Answer 13

we did not get anywhere close to integration yet, but yes, when you integrate 2 from 1 to 2 you will get 2

we still have to find the antiderivative of the second part, and for this we need "partial fractions"

Answer 14

\[\frac{-x^2+28}{x^2(x+2)}=\frac{a}{x}+\frac{b}{x^2}+\frac{c}{x+2}\] and now we need a, b and c

Answer 15

do you know how to do the partial fraction decomposition?

Answer 16

nope i am new to this, what is the answer finally?

Answer 17

for the partial fraction you will get
\[\frac{-x^2+28}{x^2(x+2)}=\frac{14}{x}-\frac{7}{x^2}+\frac{6}{x+2}\]

Answer 18

so in total
\[\frac{2x^3+3x^2+28}{x^3+2x^2}=2+\frac{14}{x}-\frac{7}{x^2}+\frac{6}{x+2}\]

Answer 19

now you have to integrate term by term to find the anti derivative

Answer 20

you get
\[\int 2+\frac{14}{x}-\frac{7}{x^2}+\frac{6}{x+2} dx = 2x+14\ln(x)+\frac{7}{x}+6\ln(x+2)\]

Answer 21

thats is not the hard part, that you just take the anti derivative term by term. the hard part is finding that
\[\frac{-x^2+28}{x^2(x+2)}=\frac{14}{x}-\frac{7}{x^2}+\frac{6}{x+2}\]

Answer 22

hm ok bro thanks..