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

[SclayP]

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.\int 1/((1+x^2)*(x+1))\,dx
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

 

[jbunniii]

Have you tried splitting the integrand into partial fractions?

 

[SclayP]

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.

 

[jbunniii]

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:
$$\frac{1}{(x^2 + 1)((x+1)} = \frac{?}{x^2 + 1} + \frac{?}{x + 1}$$

 

[jbunniii]

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.

 

[SclayP]

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:
$$\frac{1}{(x^2 + 1)((x+1)} = \frac{?}{x^2 + 1} + \frac{?}{x + 1}$$

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:

\frac{A}{x^2+1} + \frac{B}{x+1} = \frac{1}{(x^2+1)(x+1)}

Then...

A(x+1) + B(x^2+1) = 1

So,

Ax + A + Bx^2 + B = 1

So now i do this,

Ax = 0
Bx^2 = 0
A + B = 1

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

Thanks again.

 

[Dick]

This i how it goes:

\frac{A}{x^2+1} + \frac{B}{x+1} = \frac{1}{(x^2+1)(x+1)}

Then...

A(x+1) + B(x^2+1) = 1

So,

Ax + A + Bx^2 + B = 1

So now i do this,

Ax = 0
Bx^2 = 0
A + B = 1

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.

 

[SclayP]

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:

(Ax+B)(x+1) + C(x^2+1) = 1

→ Ax^2 + Ax + Bx + B + Cx^2 + C = 1
→ x^2(A +C) + x(A +B) + B + C = 1

A + C = 0
A + B = 0
B + C = 1

So...

A = -C
A = -B

Then...

-C = -B → B = C

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

 

[Dick]

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

(Ax+B)(x+1) + C(x^2+1) = 1

→ Ax^2 + Ax + Bx + B + Cx^2 + C = 1
→ x^2(A +C) + x(A +B) + B + C = 1

A + C = 0
A + B = 0
B + C = 1

So...

A = -C
A = -B

Then...

-C = -B → B = C

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

 

[Mark44]

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).

 

[Dick]

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.