Resolving Integrals by Parts: Why Can't I Do It?

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In summary, the conversation is about a person trying to solve an integral using integration by parts and coming up with a solution, but then realizing it can also be solved by partial fractions. However, they are struggling with the partial fractions method and seeking help in understanding the process. The experts explain that more variables are needed for factors with powers greater than one and provide a resource for further understanding.
  • #1
SclayP
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1. So, I have this integral which i'll post in a moment. The thing is that i attempted to resolve the integral by parts and i came up with a solution, but then i looked up in the internet and the way it's resolved it's very very different. I searched it on wolfram and i even have the step by step resolution but i don't know why i can't do it by parts. In Wolfram it's resolved first of all by "simple fractions"??
2.[itex]\int 1/((1+x^2)*(x+1))\,dx[/itex]
3. Well, like i said i tried to resolve it by part choosing 1/(x^2+1) dx like dv and 1/(x+1) like u. The i integrated by part and after all the precidure i came up with the next solution; I=arctg(x)/(x+1)

Please i'd be really happy if you could help me with this. AND SORRY FOR MY ENGLISH
 
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  • #2


Have you tried splitting the integrand into partial fractions?
 
  • #3


jbunniii said:
Have you tried splitting the integrand into partial fractions?

Yes, I've tried but i don't come up with nothing that way. I mean, I'm clearly splitting the integrand wrong, but...can't it be resolved by parts. Why ?

Thank you.
 
  • #4


SclayP said:
Yes, I've tried but i don't come up with nothing that way. I mean, I'm clearly splitting the integrand wrong, but...can't it be resolved by parts. Why ?
Integration by parts doesn't necessarily work for all integrands, even those that are a product of two functions. Over time you learn by experience what is likely to work in what case. Please show what you got when you tried partial fractions:
$$\frac{1}{(x^2 + 1)((x+1)} = \frac{?}{x^2 + 1} + \frac{?}{x + 1}$$
 
  • #5


P.S. Are you saying that you used integration by parts and got the right answer? If so, there is nothing wrong with that. Often there is more than one way to solve a problem. If you post your work I will check it for you.
 
  • #6


jbunniii said:
Integration by parts doesn't necessarily work for all integrands, even those that are a product of two functions. Over time you learn by experience what is likely to work in what case. Please show what you got when you tried partial fractions:
$$\frac{1}{(x^2 + 1)((x+1)} = \frac{?}{x^2 + 1} + \frac{?}{x + 1}$$

jbunniii said:
P.S. Are you saying that you used integration by parts and got the right answer? If so, there is nothing wrong with that. Often there is more than one way to solve a problem. If you post your work I will check it for you.

This i how it goes:

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

Then...

[itex]A(x+1) + B(x^2+1) = 1[/itex]

So,

[itex]Ax + A + Bx^2 + B = 1[/itex]

So now i do this,

[itex]Ax = 0[/itex]
[itex]Bx^2 = 0[/itex]
[itex]A + B = 1[/itex]

And this is where i don't know how to proceed...

Thanks again.
 
  • #7


SclayP said:
This i how it goes:

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

Then...

[itex]A(x+1) + B(x^2+1) = 1[/itex]

So,

[itex]Ax + A + Bx^2 + B = 1[/itex]

So now i do this,

[itex]Ax = 0[/itex]
[itex]Bx^2 = 0[/itex]
[itex]A + B = 1[/itex]

And this is where i don't know how to proceed...

Thanks again.

Actually it goes ##\frac{Ax+B}{x^2+1} + \frac{C}{x+1} = \frac{1}{(x^2+1)(x+1)}##. Try it starting from there.
 
  • #8


Dick said:
Actually it goes ##\frac{Ax+B}{x^2+1} + \frac{C}{x+1} = \frac{1}{(x^2+1)(x+1)}##. Try it starting from there.

I've tried it that way and i come up with this:

[itex](Ax+B)(x+1) + C(x^2+1) = 1[/itex]

→ [itex]Ax^2 + Ax + Bx + B + Cx^2 + C = 1[/itex]
→ [itex]x^2(A +C) + x(A +B) + B + C = 1[/itex]

[itex]A + C = 0[/itex]
[itex]A + B = 0[/itex]
[itex]B + C = 1[/itex]

So...

[itex]A = -C[/itex]
[itex]A = -B[/itex]

Then...

[itex]-C = -B[/itex] → [itex]B = C[/itex]

So, the only way is that B = 1/2 and C = 1/2

I guess its ok, but i have a doubt. How did you know it was that way, i mean i would never had though it that way. Why the x multipying the A ? And well would you explain please that ?

Thanks
 
  • #9


SclayP said:
I've tried it that way and i come up with this:

[itex](Ax+B)(x+1) + C(x^2+1) = 1[/itex]

→ [itex]Ax^2 + Ax + Bx + B + Cx^2 + C = 1[/itex]
→ [itex]x^2(A +C) + x(A +B) + B + C = 1[/itex]

[itex]A + C = 0[/itex]
[itex]A + B = 0[/itex]
[itex]B + C = 1[/itex]

So...

[itex]A = -C[/itex]
[itex]A = -B[/itex]

Then...

[itex]-C = -B[/itex] → [itex]B = C[/itex]

So, the only way is that B = 1/2 and C = 1/2

I guess its ok, but i have a doubt. How did you know it was that way, i mean i would never had though it that way. Why the x multipying the A ? And well would you explain please that ?

Thanks

You generally need more variables than one if the factor has power greater than one. If you don't have enough variables you won't find any solution to the partial fractions problem - as you noticed. If the factor is quadratic, like x^2+1 you need a linear function having two variables (A,B) in the numerator, like Ax+B. The same sort of thing happens if you have powers of polynomials, like (x+1)^2. See http://tutorial.math.lamar.edu/Classes/CalcII/PartialFractions.aspx
 
  • #10


Dick said:
You generally need more variables than one if the factor has power greater than one. If you don't have enough variables you won't find any solution to the partial fractions problem - as you noticed. If the factor is quadratic, like x^2+1 you need a linear function having two variables (A,B) in the numerator, like Ax+B. The same sort of thing happens if you have powers of polynomials, like (x+1)^2. See http://tutorial.math.lamar.edu/Classes/CalcII/PartialFractions.aspx
Just to expand on what Dick said
If the factor is an irreducible quadratic, like x^2+1, then ...

If the quadratic is factorable, then you factor it and write expressions such as A/(x - r) + B/(x - s).
 
  • #11


Mark44 said:
Just to expand on what Dick said
If the factor is an irreducible quadratic, like x^2+1, then ...

If the quadratic is factorable, then you factor it and write expressions such as A/(x - r) + B/(x - s).

You CAN do the partial fractions formalism on reducible polynomials. E.g. (3x^2+12x+11)/((x^2+3x+2)(x+3)) can be split into (2x+3)/(x^2+3x+2)+1/(x+3). But, of course, you would be a lot better off if you recognize (x^2+3x+2)=(x+1)(x+2) and split it into 1/(x+1)+1/(x+2)+1/(x+3). It's just a lot simpler.
 

What is an integral?

An integral is a mathematical concept that represents the area under a curve in a graph. It is used to find the total amount of something that is continuously changing, such as the distance traveled by a moving object or the amount of water flowing in a river.

What is the process for solving an integral?

The process for solving an integral involves finding the antiderivative of the function being integrated, setting up the limits of integration, and evaluating the integral using the fundamental theorem of calculus.

What types of integrals are there?

There are two main types of integrals: definite and indefinite. A definite integral has specific limits of integration and gives a numerical value, while an indefinite integral has no limits and gives a general expression with a constant of integration.

What are some common techniques for solving integrals?

Some common techniques for solving integrals include substitution, integration by parts, partial fractions, and trigonometric substitutions. It is important to determine the most appropriate technique for a given integral based on the form of the integrand.

Why is it important to be able to solve integrals?

Integrals are used in various fields of science, engineering, and economics to solve real-world problems. They also have important applications in physics, such as calculating the work done by a force or the average speed of an object. Being able to solve integrals is essential for understanding and solving many practical problems.

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