# Integration of Inverse of f(x)

1. ### JayKo

128
[SOLVED] Integration of Inverse of f(x)

$$\int\frac{1}{y^{4}-6y^{3}+5y^{2}}$$ dy =

$$\frac{1}{4y^{3}-18y^{2}+10y}$$(ln|$${y^{4}-6y^{3}+5y^{2}}$$|)

is this correct way or doing it? thanks. i was a bit of confuse, if i should do it the above or partial fraction the denominator it? thanks

2. ### mjsd

860
yes, factorise and then partial fractions is ok

3. ### JayKo

128
wait a minutes,i show my partial fraction. please point out where gone wrong ;) thanks

4. ### mjsd

860
firstly you can further factorise y^2-6y+5

5. ### JayKo

128
$$\frac{1}{y^{4}-6y^{3}+5y^{2}}$$

=$$\frac{1}{y^{2}(y^{2}-6y+5)}$$

=$$\frac{1}{y^{2}(y-1)(y-5)}$$

So, by Partial fraction, is this correct?

$$\frac{1}{y^{2}(y-1)(y-5)}$$
=$$\frac{A}{y}$$+$$\frac{B}{y^{2}}$$+$$\frac{C}{y-1}$$+$$\frac{D}{y-5}$$

and then, by comparing coefficient of y^3, y^2, y.

(A+C+D){y^{3}}=0
(-6A+B-5C-D){y^{2}}=0
(5A-6B)y=0
answer is A=6/25, B=1/5, C=3/10. D=-87/50.

but doubt my answer. anyone can point out?thanks

Last edited: Jan 13, 2008
6. ### JayKo

128
i know, is just that i m slow using teX, please gimme a min.thanks

7. ### mjsd

860
I can tell you the answer....

no need use latex for such simple expression

most likely you have treated the y^2 bit incorrectly

Last edited: Jan 13, 2008
8. ### JayKo

128
after i substitute A,B,C,and D.

i can get back my original 1/f(x)

9. ### mjsd

860
looks ok so far

10. ### JayKo

128
lol, i uses teX as its more tidy, i go n check the y^2 bit now.thanks.

860
FYI i've got
A=6/25
B=1/5
C=-1/4
D=1/100

12. ### JayKo

128
if so, then
$$\frac{1}{y^{2}(y-1)(y-5)}$$=$$\frac{6}{25y}$$+$$\frac{1}{5y^{2}}$$+$$\frac{3}{10(y-1)}$$-$$\frac{87}{50(y-5}$$

Last edited: Jan 13, 2008
13. ### JayKo

128
OOops,gimme a minute,i recheck :)

14. ### JayKo

128
you are right! :D :D :D thanks a lot.for pointing it out !:D

15. ### JayKo

128
$$\int\frac{1}{y^{4}-6y^{3}+5y^{2}}$$ dy =

6/25(ln|y|)-1/4(ln|y-1|)+1/100(ln|y-5|)-1/(5y). right?

but this problem is actually make y the subject in this equation, how to go about it further?
$$\int\frac{1}{y^{4}-6y^{3}+5y^{2}}$$ dy = $$\int\$$ dt

the original question is $$\frac{dy}{dt}$$=$$y^{4}-6y^{3}+5y^{2}$$ find y(t).

if $$\int\frac{1}{y^{4}-6y^{3}+5y^{2}}$$ dy =

6/25(ln|y|)-1/4(ln|y-1|)+1/100(ln|y-5|)-1/(5y)
then i m stuck to make y the subject. anyone to point out the mistake.thanks

Last edited: Jan 13, 2008
16. ### JayKo

128
anyone, can show me the right way?

17. ### JayKo

128
another differential eq. question.

y'=1+xy find y. how to go about it. i m clueless
my method of multiply dx to (1+xy) doesn't work in this case.

18. ### rohanprabhu

413
You can solve this by the solution for Linear Differential equation. The equation you have is:

$$\frac{dy}{dx} = 1 + xy$$

which can be written as:

$$\frac{dy}{dx} + y(-x) = 1$$

which is similar to:

$$\frac{dy}{dx} + yP = Q$$

Here, $P = (-x); Q = 1$

So, you have:

$$I.F = e^{\int (-x)dx}$$
$$I.F = e^{-\frac{x^2}{2}}$$

And hence, your solution is given by:

$$ye^{\frac{-x^2}{2}} = \int (1)e^{\frac{-x^2}{2}}dx + C$$

But, it's gonna be a real pain with this:

$$\int e^{\frac{-x^2}{2}}dx$$

a method to which i can't think of right now. Maybe there's some other way..

Last edited: Jan 13, 2008
19. ### Gib Z

3,348
That integral evaluates to the error function. No "method" for that one, you just have to know it. Haven't checked the rest of your post though, so not sure if the rest of your solution is right.

20. ### JayKo

128
great, this is something new to me, just started my linear algebra course.thanks.will study your solution