I need some guidance with this integration

[James10492]

Hi there, I am stuck on this integration problem and would be very appreciative of some guidance. I have tried a few things now and my methods do not seem to reach a satisfactory conclusion. I think the solution involves some sort of combination of solving by substitution and integration by parts but apparently I cannot work out the right method. Here is the problem and what I have tried:

Find

$$\int x^3 \ln {(x^2 +1)} \ dx$$
let's start with the substitution method,

$$u = x^2+1$$
now you can re-write the problem:

$$\int\frac{(u-1)}{2}\ln{u} \:du$$
$$\frac{1}{2}\int(u-1)\ln{u} \:du$$
which looks like it could be amenable to the method of integration by parts:

$$u = u-1 \\du = 1$$
$$v = u \ln{u} - u \\ dv = \ln{u}$$
$$I = (u-1)(u\ln{u} - u) - \int u\ln{u} - u$$
which gives rise to a successive integration by parts, and the problem becomes circular.

I think this is more complicated than it should be, is there something else I should try?

 

[topsquark]

Hi there, I am stuck on this integration problem and would be very appreciative of some guidance. I have tried a few things now and my methods do not seem to reach a satisfactory conclusion. I think the solution involves some sort of combination of solving by substitution and integration by parts but apparently I cannot work out the right method. Here is the problem and what I have tried:

Find

$$\int x^3 \ln {(x^2 +1)} \ dx$$
let's start with the substitution method,

$$u = x^2+1$$
now you can re-write the problem:

$$\int\frac{(u-1)}{2}\ln{u} \:du$$
$$\frac{1}{2}\int(u-1)\ln{u} \:du$$
which looks like it could be amenable to the method of integration by parts:

$$u = u-1 \\du = 1$$
$$v = u \ln{u} - u \\ dv = \ln{u}$$
$$I = (u-1)(u\ln{u} - u) - \int u\ln{u} - u$$
which gives rise to a successive integration by parts, and the problem becomes circular.

I think this is more complicated than it should be, is there something else I should try?

Try this. Integrate by parts on the original integrand:
$\int p ~ dq = pq - \int q dp$

and let $p = ln(x^2 + 1)$ and $dq = x^3 ~ dx$

-Dan

 

[James10492]

Try this. Integrate by parts on the original integrand:
$\int p ~ dq = pq - \int q dp$

and let $p = ln(x^2 + 1)$ and $dq = x^3 ~ dx$

-Dan

I get

$$I = \frac{x^4}{4}\ln{(x^2+1)} - \int\frac{x^5}{x^2+1}$$
and now I am not sure what to do, that integrand is rather awkward.

 

[BigBeachBanana]

I get

$$I = \frac{x^4}{4}\ln{(x^2+1)} - \int\frac{x^5}{x^2+1}$$
and now I am not sure what to do, that integrand is rather awkward.

Let $u=x^2+1$

 

[Dr.Peterson]

I get

$$I = \frac{x^4}{4}\ln{(x^2+1)} - \int\frac{x^5}{x^2+1}$$
and now I am not sure what to do, that integrand is rather awkward.

... or start by doing a long division.

 

[Steven G]

Hi there, I am stuck on this integration problem and would be very appreciative of some guidance. I have tried a few things now and my methods do not seem to reach a satisfactory conclusion. I think the solution involves some sort of combination of solving by substitution and integration by parts but apparently I cannot work out the right method. Here is the problem and what I have tried:

Find

$$\int x^3 \ln {(x^2 +1)} \ dx$$
let's start with the substitution method,

$$u = x^2+1$$
now you can re-write the problem:

$$\int\frac{(u-1)}{2}\ln{u} \:du$$
$$\frac{1}{2}\int(u-1)\ln{u} \:du$$
which looks like it could be amenable to the method of integration by parts:

$$u = u-1 \\du = 1$$
$$v = u \ln{u} - u \\ dv = \ln{u}$$
$$I = (u-1)(u\ln{u} - u) - \int u\ln{u} - u$$
which gives rise to a successive integration by parts, and the problem becomes circular.

I think this is more complicated than it should be, is there something else I should try?

Please do not use u to represent different things. Please! Also, your du does not equal 1.

 

[Steven G]

Whenever you are integrating a rational expression where the degree of the numerator is greater than or equal to the degree of the denominator then, as Dr P suggested, do long division.

 

[James10492]

Please do not use u to represent different things. Please! Also, your du does not equal 1.

Haha ok I won't do that again.

Whenever you are integrating a rational expression where the degree of the numerator is greater than or equal to the degree of the denominator then, as Dr P suggested, do long division.

I got it now, didn't think to try that before. You guys rock!

One thing though, I tried to integrate the x^5/(x^2+1) term using the substitution method and arrived at a different (wrong) solution... why can't it be solved this way? If the degree of the numerator is higher than the denominator do you always have to analyse a problem like this?

 

[Dr.Peterson]

One thing though, I tried to integrate the x^5/(x^2+1) term using the substitution method and arrived at a different (wrong) solution... why can't it be solved this way? If the degree of the numerator is higher than the denominator do you always have to analyse a problem like this?

Please show that work, so we can see what went wrong. It is possible to do it that way, and I get the same answer both ways; you may have made an error in doing it, or you don't see that the answers are equal.

(Also show the "correct" answer you get, so we can make sure that's right!)

 

[lex]

I get

$$I = \frac{x^4}{4}\ln{(x^2+1)} - \int\frac{x^5}{x^2+1}$$

There is a $\tfrac{1}{2}$ missing in front of this integral.
$$I = \frac{x^4}{4}\ln{(x^2+1)} - \tfrac{1}{2}\int\frac{x^5}{x^2+1}\,dx$$

 

[Steven G]

There is a dx that the OP also left out.

 

[lex]

There is a dx that the OP also left out.

Yes!