$\displaystyle
\int\frac{1}{x^2+x-2}=\frac{ln(x^2+x-2)}{2x+1}
$
is this right?
$\displaystyle \int\frac{1}{x^2+x-2} \ dx=\frac{1}{(x+2)(x-1)}$
Use partial fractions:
$\displaystyle \frac{1}{(x+2)(x-1)}=\frac{A}{x+2}+\frac{B}{x-1}$
$\displaystyle A(x-1)+B(x+2)=1$
$\displaystyle A=-\frac{1}{3}, B=\frac{1}{3}$
$\displaystyle \int\frac{1}{x^2+x-2} \ dx= \int\frac{-\frac{1}{3}}{x+2}+\frac{\frac{1}{3}}{x-1} \ dx$
$\displaystyle =\frac{1}{3}\ln (x-1)-\frac{1}{3}\ln (x+2) +C$
$\displaystyle \displaystyle\frac{A}{x+2}+\frac{B}{x-1}=\frac{1}{(x+2)(x-1)}$
$\displaystyle A(x-1)+B(x+2)=1$
$\displaystyle A=-\frac{1}{3}$
$\displaystyle B=\frac{1}{3}$
$\displaystyle \displaystyle\int\frac{\frac{1}{3}}{(x-1)}-\frac{\frac{1}{3}}{(x+2)}dx $$\displaystyle \displaystyle\implies$
$\displaystyle \displaystyle\implies\frac{1}{3}\int\frac{1}{x-1}-\frac{1}{x+2}=\frac{1}{3}(ln|x-1|-ln|x+2|)=\frac{1}{3}ln|\frac{x-1}{x+2}|+C$
Hope that this help
No, it is not. When you are differentiating and using the chain rule you can multiply by the derivative:
If f(x)= f(u(x)) then $\displaystyle df/dx= f'(u(x)) u'(x)$
but that leads to the integration rule $\displaystyle \int f'(u(x)) u'(x)dx$ we you can substitute for u(x) to get $\displaystyle \int f'(u)du$ and get f(u)+ C= f(u(x))+ C.
But if you do NOT already have that deriviative inside the integral, you cannot put it in. That is, you cannot say
$\displaystyle \displaystyle \int f(u(x))dx= \frac{f'(u)}{f'(u)}\int f(u)dx= \frac{1}{f'(u)}\int f(u) f'(u)dx= \frac{1}{f'(u}}\int f(u)du$
because you can NOT move that function of x, f'(u(x)), inside the integral sign like that.
The standard method of integrating such a "rational function" is to use "partial fractions", arcevipa shows. Notice that you can do that with linear terms, such as "mx+ b" because the derivative is a constant and you can move constants inside and outside of the integral:
$\displaystyle \displaystyle \int \frac{1}{mx+ b}dx= \frac{m}{m}\int\frac{1}{mx+ b} dx= \frac{1}{m}\int\frac{1}{mx+ b}(m dx)$ and now you can let u= mx+ b so that becomes $\displaystyle \displaystyle \frac{1}{m}\int \frac{1}{u}du= \frac{1}{m} ln(u)+ C= \frac{1}{m}ln(mx+ b)+ C$.
But you cannot do that with functions of x: $\displaystyle c\int g(x)dx= \int cg(x)dx$ for c a constant, but $\displaystyle f(x)\int g(x)dx\ne \int f(x)g(x)dx$ with the function f(x).
Or to double avoid partial fractions $\displaystyle ($a habit that I'm picking-up from MHF integral masters $\displaystyle )$:
Make the following substitution:
$\displaystyle \displaystyle t = \frac{x-1}{x+2}$ $\displaystyle \displaystyle
\Rightarrow \dfrac{dt}{dx} = \frac{3}{(x+2)^2}$ $\displaystyle \displaystyle \Rightarrow dx = \dfrac{(x+2)^2}{3}\;{dt}$
So $\displaystyle \displaystyle \int\frac{1}{x^2+x-2}\;{dx} = \int\frac{1}{(x+2)(x-1)}\;{dx} = \frac{1}{3}\int\frac{(x+2)^2}{(x+2)(x-1)}\;{dt}$ $\displaystyle
\displaystyle =\frac{1}{3}\int\frac{x+2}{x-1}\;{dt} $
$\displaystyle \displaystyle t = \frac{x-1}{x+2}$ $\displaystyle \displaystyle
\Rightarrow t(x+2) = (x-1)$, so we have:
$\displaystyle \displaystyle \frac{1}{3}\int\frac{x+2}{x-1}\;{dt} = \frac{1}{3}\int\frac{x+2}{t(x+2)}\;{dt} = \frac{1}{3}\int\frac{1}{t}\;{dt} = \frac{1}{3}\ln{t}+k$
Therefore $\displaystyle \displaystyle \int\frac{1}{x^2+x-2}\;{dx} = \frac{1}{3}\ln\left(\frac{x-1}{x+2}\right)+k$.
Thanks guys, from what I understand, partial fractions make the constants so that we can remove constants from the integral.
Now I have a question.. what if a partial fraction is in the form of for example $\displaystyle \frac{5x}{x^2-1}$?
Can I do this?
$\displaystyle \int\frac{5x}{x^2-1}=5x\int\frac{1}{x^2-1}$
$\displaystyle =\frac{5xln(x^2-1)}{2x}$
First, that is NOT "a partial fraction" "in the form $\displaystyle \frac{5x}{x^2- 1}$", it is simply a fraction. The "partial fractions" are what you get when you separate it into a sum of simpler fractions.
Second, you were just told that you cannot do that! And you just said "make the constants so we can remove constants from the integral". Why are you now asking essentially the same question you asked before? You cannot just take the "5x" out in front of the integral and you cannot then just divide by 2x after integrating. You cannot do that with any functions of x- only with constants.
(If you were going to factor 5x out of the integral, why not just factor out the whole thing? Would you claim that $\displaystyle \int\frac{5x}{x^2- 1}= \frac{5x}{x^2- 1}\int 1 dx= \frac{5x}{x^2- 1}+ C$? Would you claim that $\displaystyle \int f(x)dx= f(x)\int dx= f(x)+ C$. That would make integration very easy! (Indeed, it would make it so trivial, it would be useless!)
$\displaystyle x^2- 1= (x- 1)(x+ 1)$ so $\displaystyle \frac{5x}{x^2- 1}= \frac{A}{x- 1}+ \frac{B}{x+ 1}$ for some numbers, A and B.
If you multiply both sides by (x- 1)(x+ 1), you get 5x= A(x+ 1)+ B(x- 1). Now there are several ways to determine A and B
1) Multiply the right side out so 5x= Ax+ A+ Bx- B= (A+ B)x+ (A- B). If two polynomials are equal for all x, they must have the same coefficients so A+ B= 5, A- B= 0. That gives you two equations to solve for A and B.
2) Choose any two values of x you like to get two equations. If you choose x= 2, then 5x= A(x+ 1)+ B(x- 1) becomes 10= 3A+ B. If you choose x= 3, then 5x= A(x+ 1)+ B(x- 1) becomes 15= 4A+ 2B.
3) Often (but not always) you can choose numbers to immediately eliminate one of the unknown coefficients. If x= 1 then x- 1= 0 so taking x= 1 gives 5= A(1+ 1)+ B(1- 1)= 2A. If x=-1 then x+1= 0 so taking x= -1 gives -5= A(1- 1)+ B(-1- 1)= -2B.
By the way, in writing out the integrals you did not include "dx" and you did not include the constant of integration "+ C". A good teacher would mark down for both of those.
No, you won't get that.
$\displaystyle \displaystyle\frac{5x}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}$
$\displaystyle \displaystyle\frac{x+1}{x+1}\ \frac{A}{x-1}+\frac{x-1}{x-1}\ \frac{B}{x+1}=\frac{5x}{(x-1)(x+1)}$
$\displaystyle \Rightarrow\ A(x+1)+B(x-1)=5x\Rightarrow\ A+B=5,\;\;A-B=0\Rightarrow\ A=B$
Hold on.. I think all of u are mistaken
What I meant : I am given a function F(x)
I used partial fractions on F(x) and found that F(x)= A + B +$\displaystyle \int\frac{5x}{x^2-1}dx$
I then am required to integrate F(x)
I have no problem with integrating A and B but faced a problem with $\displaystyle \int\frac{5x}{x^2-1}dx$
so my question is... if $\displaystyle \int\frac{5x}{x^2-1}dx$ is one of the partial fractions, can I integrate it this way?
$\displaystyle \int\frac{5x}{x^2-1}dx=5x\int\frac{1}{x^2-1}dx$
$\displaystyle =\frac{5xln(x^2-1)}{2x}$
No, you cannot take the variable x outside the integral.
If you have A+B+integral, split the remaining fraction that you are integrating into partial fractions.
Use C and D for numerators if you like.
A simplification is made by rewriting the fraction you need to integrate as a pair of simpler fractions.
You can easily integrate the resulting partial fractions.
When you add those fractions together, you get the given fraction.
Hence they are the same but it's far simpler to integrate the partial fractions seperately.
If that integral is part of your problem, can you post the full problem?
For the third time: NO, you cannot integrate $\displaystyle \int\frac{5x}{x^2-1}dx$ that way. What part of the earlier posts made this unclear? Please try to learn from the replies you were given, especially since some of them have quite obviously invested a great deal of time in trying to explain things to you.
For the record, you calculate $\displaystyle \int\frac{5x}{x^2-1}dx$ by using the substitution u = x^2 - 1.