# Iterated integral

#### Petrus

##### Well-known member
Hello MHB,

$$\displaystyle \int_4^1\int_1^2 \frac{x}{y}+\frac{y}{x}dydx$$
My progress:
$$\displaystyle \int_1^4[x\ln(y)+\frac{y^2}{2x}]_1^2$$
$$\displaystyle \int_1^4 x\ln(2)+\frac{2}{x}-\frac{1}{2x}dx$$
$$\displaystyle [\frac{x^2\ln(2)}{2}+2\ln(2)-\ln(2x)]_1^4$$
What I am doing wrong?

Regards,

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: iterated integral

Hello MHB,

$$\displaystyle \int_4^1\int_1^2 \frac{x}{y}+\frac{y}{x}dydx$$
My progress:
$$\displaystyle \int_1^4[x\ln(y)+\frac{y^2}{2x}]_1^2$$
$$\displaystyle \int_1^4 x\ln(2)+\frac{2}{x}-\frac{1}{2x}dx$$
$$\displaystyle [\frac{x^2\ln(2)}{2}+2\ln(2)-\ln(2x)]_1^4$$
What I am doing wrong?

Regards,

Edit: Can't change subject, I did misspelled 'Iterated integral'
How did you integrate $$\displaystyle \frac {1}{2x}$$ ?

#### Petrus

##### Well-known member
Re: iterated integral

How did you integrate $$\displaystyle \frac {1}{2x}$$ ?
Typo should be $$\displaystyle \frac{ln(2x)}{2}$$

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Ohh I see..... I antiderivated it wrong.... Should be $$\displaystyle \frac{ln(x)}{2}$$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: iterated integral

Typo should be $$\displaystyle \frac{ln(2x)}{2}$$

- - - Updated - - -

Ohh I see..... I antiderivated it wrong.... Should be $$\displaystyle \frac{ln(x)}{2}$$
They are actually the same , can you see why ?

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: iterated integral

$$\displaystyle [\frac{x^2\ln(2)}{2}+2\ln(2)-\ln(2x)]_1^4$$
Also you should have got $$\displaystyle 2\ln(x)$$ not $$\displaystyle 2\ln (2)$$

#### Petrus

##### Well-known member
Re: iterated integral

They are actually the same , can you see why ?
I think I need tea.. Yeah I see they are same
$$\displaystyle \frac{ln(2x)}{2}$$ if we derivate we get
$$\displaystyle \frac{\frac{2}{2x}}{2}<=>\frac{1}{2x}$$
$$\displaystyle \frac{ln(x)}{2}$$ if we derivate we get $$\displaystyle \frac{\frac{1}{x}}{2} <=> \frac{1}{2x}$$

#### Petrus

##### Well-known member
Re: iterated integral

Do I got more misstake that I can't see cause I get the answer and facit says $$\displaystyle \frac{21}{2}\ln(2)$$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: iterated integral

This is related to the constant of integration

$$\displaystyle \int \frac{1}{2x} dx = \frac { \ln (2x) }{2} + c_1 = \frac { \ln (2) +\ln (x) }{2} + c_1$$

Now assume that $$\displaystyle c= \frac { \ln (2) }{2} + c_1$$

Hence

$$\displaystyle \int \frac{1}{2x} dx = \frac { \ln (x) }{2} + c$$

#### Petrus

##### Well-known member
Re: iterated integral

This is related to the constant of integration

$$\displaystyle \int \frac{1}{2x} dx = \frac { \ln (2x) }{2} + c_1 = \frac { \ln (2) +\ln (x) }{2} + c_1$$

Now assume that $$\displaystyle c= \frac { \ln (2) }{2} + c_1$$

Hence

$$\displaystyle \int \frac{1}{2x} dx = \frac { \ln (x) }{2} + c$$
Thanks!But have I antiderivated correct?

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: iterated integral

Do I got more misstake that I can't see cause I get the answer and facit says $$\displaystyle \frac{21}{2}\ln(2)$$
Try evaluating it by hand , it should be the same .