Question

Find the partial fractions decomposition and an antiderivative.

Answer 1

\[\frac{ 2x-3 }{ (x+2)^{2} }= \frac{ 2x-3 }{ (x+2)(x+2) }\]using this you can break down the fraction by readjusting the denominator: \[\frac{ 2x-3 }{ (x+2)(x+2) }= \frac{ A }{ (x+2) }+\frac{ B }{ (x+2) }\] can you take it from there?

Answer 2

do i do cross multiply next?

Answer 3

Yep. you should get a linear equation such that: \[2x - 3=A(x+2)+B(x+2)\]

Answer 4

okay so what do i do next?

Answer 5

You're going to evaluate A and B with respect to the constants and coefficients with the variables. By that I mean you're going to solve for A and for B. Let me know once you've attempted this and I will post further :)

Answer 6

ok

Answer 7

okay so i picked -2 as x so 2 can become 0 and i'm going to substitute into the problem to solve.

Answer 8

\[\frac{ 2x-3 }{\left( x+2 \right)^{2} }=\frac{ 2x+4-4-3 }{ \left( x+2 \right)^{2} }\]
\[\frac{ 2x+4-7 }{\left( x+2 \right)^{2} }=\frac{ 2x+4 }{ \left( x+2 \right)^{2} }-\frac{ 7 }{ \left( x+2 \right)^{2} }\]
\[=\frac{ 2\left( x+2 \right) }{ \left( x+2 \right)^{2} }-\frac{7 }{\left( x+2 \right)^{2} }\]
\[=\frac{ 2 }{x+2 }-\frac{ 7 }{ \left( x+2 \right)^{2} }\]

Answer 9

She wants to solve it using partial fractions... And @onegirl, I will be back in a couple mins.

Answer 10

okay

Answer 11

Alright

Answer 12

so i put -2 and i got -7 = a and b :/

Answer 13

so?

Answer 14

Working on it :)

Answer 15

ok

Answer 16

Ouch.. I think I might have done it wrong :\ reworking the problem!

Answer 17

\[\int\limits \frac{ 2x-3 }{\left( x+2 \right)^{2} }dx=\int\limits \frac{ 2dx }{x+2 }-7\int\limits \left( x+2 \right)^{-2}dx
\[=2\ln \left( x+2 \right)-7\frac{ \left( x+2 \right)^{-2+1} }{-2+1}+c\] \]
\[=2\ln \left( x+2 \right)+\frac{ 7 }{x+2 }+c\]

Answer 18

ummm....thats not using partial fractions decomposition..

Answer 19

os is that the antiderivative

Answer 20

yes, i have integrated after making partial fractions.

Answer 21

ok

Answer 22

so tahts my anti derivative

Answer 23

\[\frac{ 2x-3 }{ (x+2)^{2} } = \frac{ A }{ (x+2) }+\frac{ B }{ (x+2)^{2} }\] then you have.. \[2x-3 = A(x+2) +B\] Evaluating for A and B I get: \[A= 2, B=-3\] Putting it in Integral form,I get :\[\int\limits \frac{ 2 }{ (x+2) }-\frac{ 3 }{ (x+2)^{2} } dx\] \[2 \int\limits \frac{ 1 }{ (x+2) } - 3 \int\limits \frac{ 1 }{ (x+2)^{2} }\]\[2\ln |x+2| - 3\int\limits \frac{ 1 }{ (x+2)^{2} }\]

Answer 24

So if you've gotten this far, good. butin order to evaluate \[\int\limits \frac{ 1 }{ (x+2)^{2} } dx\] you need to use u-substitution. So!..let \[u=x+2\]\[du=dx\] and so \[\int\limits \frac{ 1 }{ (x+2)^{2} }dx = \int\limits \frac{ 1 }{ u^2}du\] That becomes \[\int\limits u ^{-2}du =-\frac{ 1 }{ u }\]

Answer 25

Altogether, we have, \[2\ln|x+2|+3\ln(x+2)^2 +C\]

Answer 26

I forgot my dx somewhere in there... :\ But you get the point!! :)

Answer 27

so thats the antiderivative

Answer 28

I believe so, yes. I don't know where @surjithayer got the -7 from.. been trying to work that out into my problem.

Answer 29

Attempted the problem, tries to show step by step working it out. Correct me if im wrong!! :)

Answer 30

\[\frac{ 2 }{ x+2 }-\frac{ 3 }{\left( x+2 \right)^{2} }=\frac{ 2\left( x+2 \right)-3 }{\left( x+2 \right)^{2} }\]
\[=\frac{ 2x+4-3 }{\left( x+2 \right)^{2} }=\frac{ 2x+1 }{\left( x+2 \right)^{2} }\neq \frac{ 2x-3 }{\left( x+2 \right)^{2} }\]

Answer 31

so i still havent gotten the partial fractions..

Answer 32

\[\frac{ 2x-3 }{\left( x+2 \right)^{2} }=\frac{ A }{x+2 }+\frac{ B }{ \left( x+2 \right)^{2}}\]
2x-3=A(x+2)+B
now you can find the values of A & B

Answer 33

The partial fraction is obtained when you break down your problem into a simpler form in order to evaluate the numerator in conjunction with the denominator. that was the A and B thing. those ARE your partial fractions.

You then take the partial fractions and integrate them as you would your original problem. using A and B just helps to make it easier to evaluate since you can't easily evaluate the problem itself.

Answer 34

ohh ok

Answer 35

So as @surjithayer stated, \[\frac{ A }{ x+2 }+\frac{ B }{ (x+2)^2 }\] are your partial fractions. Now you just evaluate the partial fractions in order to obtain a solution, the same solution you would get if you were to evaluate your problem a different way.

Answer 36

2x-3=A(x+2)+B
comparing the co-efficients of x & constant terms
A=2
-3=2A+B=2*2+B
B=-3-4=-7

Answer 37

okay

Answer 38

Ohh you're right.It is -7. Is my work completely wrong then? :\