A+B = 0A+2B = 1B = 1/2A = -1/2\frac {-1/2}{(u+2)} + \frac {1/2}{(u+1)}

[kuskus94]

First, I apologise that I have posted this question in the wrong section. I should have posted it in the precalculus section.

Homework Statement

\int \frac {1}{u^2+3u+2} du

Homework Equations

\int \frac {1}{x} dx = ln{x}

We know that \frac {d(ln(u^2+3u+2))}{dx}
=\frac {1}{u^2+3u+2} * \frac {d(u^2+3u+2)}{dx}
= \frac {2u+3}{u^2+3u+2}

The Attempt at a Solution

To remove the 2u+3 is by multiplying \frac{1}{2u+3}In the end, i get \int \frac {1}{u^2+3u+2} du = \frac {ln(u^2+3u+2)}{2u+3}.

Is this correct?

Can I use subtitution?

\int \frac {1}{u^2+3u+2} du = \int \frac {1}{a} \frac{da}{2u+3} =\frac{1}{2u+3} \int \frac {1}{a}da

for a = u^2+3u+2
\frac {da}{du} = 2u+3
\frac {da}{2u+3} =du

 

[Saitama]

In the end, i get \int \frac {1}{u^2+3u+2} du = \frac {ln(u^2+3u+2)}{2u+3}.

Is this correct?

No, this is not correct.

A better approach would be to convert the denominator in a perfect square. For example:
x^2+x+1=(x+\frac{1}{2})^2+\frac{3}{4}
This could be then evaluated using the standard integral ∫dx/(x^2+a^2) where a is a constant.

 

[Millennial]

Another approach :
The denominator has the quadratic x^2+3x+2, which is equal to (x+2)(x+1). Now, we can use partial fractions:
\frac{1}{x^2+3x+2}=\frac{1}{(x+2)(x+1)}=\frac{A}{x+2}+\frac{B}{x+1}
Would you solve that and then integrate?

 

[lurflurf]

hint
d \log(\frac{p}{q})=\frac{q dp-p dq}{p q}

 

[kuskus94]

Another approach :
The denominator has the quadratic x^2+3x+2, which is equal to (x+2)(x+1). Now, we can use partial fractions:
\frac{1}{x^2+3x+2}=\frac{1}{(x+2)(x+1)}=\frac{A}{x+2}+\frac{B}{x+1}
Would you solve that and then integrate?

Wait, shouldn't it have to be in a form like \frac{1}{x(x-2)^2}=\frac{A}{x}+\frac{B}{(x-2)^2}+\frac{C}{x-2}

hint
d \log(\frac{p}{q})=\frac{q dp-p dq}{p q}

I still don't get it. Is the "d" coefficient or the symbol of differentiation?

 

[Mark44]

Another approach :
The denominator has the quadratic x^2+3x+2, which is equal to (x+2)(x+1). Now, we can use partial fractions:
\frac{1}{x^2+3x+2}=\frac{1}{(x+2)(x+1)}=\frac{A}{x+2}+\frac{B}{x+1}
Would you solve that and then integrate?

Wait, shouldn't it have to be in a form like \frac{1}{x(x-2)^2}=\frac{A}{x}+\frac{B}{(x-2)^2}+\frac{C}{x-2}

?
Why would it have this form. x² + 3x + 2 = (x + 2)(x + 1). Where are you getting x(x - 2)²?

I still don't get it. Is the "d" coefficient or the symbol of differentiation?

The d is part of a differential - it is not a coefficient.

 

[aralbrec]

Can I use subtitution?

\int \frac {1}{u^2+3u+2} du = \int \frac {1}{a} \frac{da}{2u+3} =\frac{1}{2u+3} \int \frac {1}{a}da

for a = u^2+3u+2
\frac {da}{du} = 2u+3
\frac {da}{2u+3} =du

You can do this but your last step is wrong. u is not independent of a, it is a function of a so you cannot take it out of the integral. You can solve for u from a = u^2+3u+2 and stick that into the last step but as you see, it will not bring you closer to a solution.

To use the ln thing you are trying to use, the derivative of the denominator should appear in the numerator of the integrand. There's no obvious way to do that in this case so you should try another method. If you see a polynomial in the denominator, factoring it is usually a good first step as someone else pointed out already.

 

[kuskus94]

?
Why would it have this form. x² + 3x + 2 = (x + 2)(x + 1). Where are you getting x(x - 2)²?

Oh, it was just an example... :)

The d is part of a differential - it is not a coefficient.

How can I put it into the form of log?

You can do this but your last step is wrong. u is not independent of a, it is a function of a so you cannot take it out of the integral. You can solve for u from a = u^2+3u+2 and stick that into the last step but as you see, it will not bring you closer to a solution.

To use the ln thing you are trying to use, the derivative of the denominator should appear in the numerator of the integrand. There's no obvious way to do that in this case so you should try another method. If you see a polynomial in the denominator, factoring it is usually a good first step as someone else pointed out already.

So, the first thing I have to do is factorizing the denominator
\int \frac {1}{u^2+3u+2} du=\int \frac {1}{(u+2)(u+1)} du
and after that, I should get something like
=\int (u+2)^{-1}(u+1)^{-1} du
then, assume that (u+2) is q and (u+1) is p
=qp-\int p dq
Is this correct?

 

[Mark44]

So, the first thing I have to do is factorizing the denominator
\int \frac {1}{u^2+3u+2} du=\int \frac {1}{(u+2)(u+1)} du
and after that, I should get something like
=\int (u+2)^{-1}(u+1)^{-1} du

The form above is no help.

then, assume that (u+2) is q and (u+1) is p
=qp-\int p dq
Is this correct?

No. Integration by parts is not the way to go. Instead, use partial fraction decomposition to write $\frac{1}{(u + 2)(u + 1)}$ as $\frac{A}{u + 2} + \frac{B}{u + 1}$. The goal here is to find constants A and B so that the two sides are identically equal.

Once you have found A and B, you can write your integral as
$$\int \frac{Adu}{u + 2} + \int \frac{Bdu}{u + 1} $$

(This assumes that you have already found A and B.) The integrals above are where the ln part comes from.

 

[kuskus94]

The form above is no help.
No. Integration by parts is not the way to go. Instead, use partial fraction decomposition to write $\frac{1}{(u + 2)(u + 1)}$ as $\frac{A}{u + 2} + \frac{B}{u + 1}$. The goal here is to find constants A and B so that the two sides are identically equal.

Once you have found A and B, you can write your integral as
$$\int \frac{Adu}{u + 2} + \int \frac{Bdu}{u + 1} $$

(This assumes that you have already found A and B.) The integrals above are where the ln part comes from.

Okay, I am having trouble on determining the A and B.

\frac {1}{(u+2)(u+1)}

\frac {A}{(u+2)} + \frac {B}{(u+1)}

\frac {A(u+1)+B(u+2)}{(u+2)(u+1)}

A(u+1)+B(u+2)=1

There are two constants that I can not determine its value.

Or is it something like this :

1 = Au+A+Bu+2B
1 = Au+Bu+A+2B
1 = u(A+B)+(A+2B)

since there is no u, it is considered its constant is 0

(A+B) = 0
A=-B

and

(A+2B) = 1
-B+2B=1
B=1
A=-1



=\int \frac{-1 du}{u + 2} + \int \frac{1 du}{u + 1}

= - ln(u+2) + ln(u+1)

 

[Mark44]

1=Au+Bu+A +2B, or
1 = (A + B)u + A + 2B

Since this equation has to be identically true (i.e., true for all values of u), then it must be the case that
A + B = 0 and
A + 2B = 1

From the first equation, A = -B

Substituting into the second equation, we have
-B + 2B = 1, so B = 1 and A = - 1

This is what you got, but I'm trying to explain what it is that you're doing.

From your work,
$$ \int \frac{du}{(u + 1)(u + 2)} = -ln(u + 2) + ln(u + 1) + C$$

How can you check to see if an antiderivative is correct?

 

[kuskus94]

1=Au+Bu+A +2B, or
1 = (A + B)u + A + 2B

Since this equation has to be identically true (i.e., true for all values of u), then it must be the case that
A + B = 0 and
A + 2B = 1

From the first equation, A = -B

Substituting into the second equation, we have
-B + 2B = 1, so B = 1 and A = - 1

This is what you got, but I'm trying to explain what it is that you're doing.



From your work,
$$ \int \frac{du}{(u + 1)(u + 2)} = -ln(u + 2) + ln(u + 1) + C$$

How can you check to see if an antiderivative is correct?

By differentiating it?


$$ \frac {d(-ln(u + 2) + ln(u + 1) + C)}{du}$$
$$ = \frac{-1}{(u+2)} + \frac {1}{(u+1)}$$
$$ = \frac {-(u+1)+(u+2)}{(u+2)(u+1)} $$
$$ = \frac {1}{(u+2)(u+1)} $$

 

[Mark44]

You answered your own question.

 

[aralbrec]

Okay, I am having trouble on determining the A and B.

\frac {1}{(u+2)(u+1)}

\frac {A}{(u+2)} + \frac {B}{(u+1)}

\frac {A(u+1)+B(u+2)}{(u+2)(u+1)}

A(u+1)+B(u+2)=1

There is a shorter way to do this. The equation has to be true for all values of u. So in your last step, what happens if u=-1 or u=-2.

You can get that straight from the expansion equation:

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

Multiply both sides by (u+2) and set u=-2. Multiply both sides by (u+1) and set u=-1.

The shortcuts are a little more difficult if the denominator is second order (eg (u²+a²) or if the roots are repeated (eg (u+a)²) but you should read the section on partial fraction expansion in your textbook to learn those cases.

In any case you can always do this:

1 = u(A+B)+(A+2B)

and equate corresonding coefficients of uⁿ as you've done.