# Log and integrals

#### mark090480

Hi,

I need some help with ln and integrals. So I am given the equation:

x^3 ln x dx

in the image below.

I have trouble with this line:

= 1/4^3 lnx - 1/4 (F) x^3

(F) being the integral sign

why is that 1/4 not in the second integral?

Hope you get what I mean.  #### Attachments

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#### harish21

Hi,

I need some help with ln and integrals. So I am given the equation:

x^3 ln x dx

in the image below.

I have trouble with this line:

= 1/4^3 lnx - 1/4 (F) x^3

(F) being the integral sign

why is that 1/4 not in the second integral?

Hope you get what I mean.  You are supposed to use integration by parts here:

let $$\displaystyle u = ln(x)$$ and $$\displaystyle dv = x^3 dx$$

then $$\displaystyle du = \frac{1}{x} dx$$ and $$\displaystyle v = \frac{x^4}{4}$$

then calculate $$\displaystyle uv - \int v \mbox{du}$$

#### mark090480

Hi,

I got that part but then shouldn't that 1/4 on the fourth line of that image be in the second integral? It is not, that is why I am stuck.

Thanks.

#### harish21

Hi,

I got that part but then shouldn't that 1/4 on the fourth line of that image be in the second integral? It is not, that is why I am stuck.

Thanks.
so you have: $$\displaystyle uv - \int v \mbox{du}$$

refer to the above post to see what u, v, du, and dv are.

Then $$\displaystyle uv - \int v \mbox{du} = ln(x) \times \frac{x^4}{4} - \int \frac{x^4}{4} \times \frac{1}{x} dx$$

$$\displaystyle = ln(x) \times \frac{x^4}{4} - \frac{1}{4}\int{x^4} \times \frac{1}{x} dx$$

Note: $$\displaystyle \int \frac{x}{4}$$ and $$\displaystyle \frac{1}{4}\int x$$ is the same.

#### 11rdc11

Hi,

I got that part but then shouldn't that 1/4 on the fourth line of that image be in the second integral? It is not, that is why I am stuck.

Thanks.

It is a constant so you can put it infront of the integral

#### mark090480

It is a constant so you can put it infront of the integral
Why is it not a coefficient? 1/4x^4 ? This is maybe why I got stuck #### HallsofIvy

MHF Helper
One of the things you should have learned when you started on integration, long before you got to "integration by parts", is that you can take constants out of integrals: $$\displaystyle \int c f(x) dx= c \int f(x)dx$$ for c any constant.

#### mark090480

I did not know it was a constant and I'm still not sure why it is.

#### Defunkt

MHF Hall of Honor
A constant is a number. $$\displaystyle \frac{1}{4}$$ is a number, therefore you can "pull it out" of the integral.

#### mark090480

I think I get it now. So using the constant multiple rule, it would make no difference if that coefficient is in the integral or not, but it could be there if we wanted and the outcome would be the same.