1. ## difficult integral

Hi everyone,

Can someone give me a hint on this one, please?

I have the integral from 1 to the square root of 3 arctan(1/x) dx

I said that u=arctan(1/x)
du=????
dv=???????
v=???????

I'm not sure if my u is correct or not. I don't think it is because now I can't choose anything for du, right? Should it be u=arctan and du=1/x????

Thank you

2. $\displaystyle u=\arctan\frac{1}{x}$
$\displaystyle du=-\frac{1}{1+x^2}$
$dv=1$
$v=x$

3. Originally Posted by chocolatelover
I have the integral from 1 to the square root of 3 arctan(1/x) dx

I said that u=arctan(1/x)
du=????
dv=???????
v=???????
First, you need to get the antiderivative of $\int\arctan\frac1x\,dx$

So you apply integration by parts.

Take $u=\arctan\frac1x$. Now, to get $du$, remember that $(\arctan y)'=\frac1{1+y^2}y'$ (chain rule); and take $dv=dx\implies v=x$

Try it, tell us later.

4. This is what I got:

u=arctan(1/x)
du=1/1+x^2dx
dv=dx
v=x

integral 1 to the square root of 3 arctan(1/x)dx=xtan^-1(1-x)]one to the square root of 3-integral 1 to the square root of 3 x/1+x^2dx=

square root of 3tan^-1(1-sqauare root of 3)-[1(tan^-1)(1-1)]-integral 1 to the square root of three x/1+x^2dx=

-1.095-0-integral 1 to the square root of three x/1+x^2dx

-1.095-the integral 0 to the square root of 9x/1+x^2dx

t=1+x^2
dt=2xdx
xdx=1/2dt

when x=square root of 3 t=4
when x=1 t=2

integral 1 to square root of 3 x/1+x^2dx=1/2 integral 2 to 4 dt/t=1/2ln(t)]2 to 4=
1/2(ln4-ln2)=1/2ln2=

integral 1 to square root of 3 tan^-1xdx=-1.095-integral 1 to square root 3 x/1+x^2=
-1.095-1/2ln2

Can you tell me if this is correct and if it isn't where I made my mistake?

Thank you very much

5. Originally Posted by chocolatelover
u=arctan(1/x)
du=1/1+x^2dx
Correct this.

6. u=arctan(1/x)
du=1/1+x^2dx

Correct this.

u=arctan(1/x)
du=[1/1+x^2]ln(x)

7. Originally Posted by chocolatelover
This is what I got:

u=arctan(1/x)
du=1/1+x^2dx
dv=dx
v=x

integral 1 to the square root of 3 arctan(1/x)dx=xtan^-1(1-x)]one to the square root of 3-integral 1 to the square root of 3 x/1+x^2dx=

square root of 3tan^-1(1-sqauare root of 3)-[1(tan^-1)(1-1)]-integral 1 to the square root of three x/1+x^2dx=

-1.095-0-integral 1 to the square root of three x/1+x^2dx

-1.095-the integral 0 to the square root of 9x/1+x^2dx

t=1+x^2
dt=2xdx
xdx=1/2dt

when x=square root of 3 t=4
when x=1 t=2

integral 1 to square root of 3 x/1+x^2dx=1/2 integral 2 to 4 dt/t=1/2ln(t)]2 to 4=
1/2(ln4-ln2)=1/2ln2=

integral 1 to square root of 3 tan^-1xdx=-1.095-integral 1 to square root 3 x/1+x^2=
-1.095-1/2ln2

Can you tell me if this is correct and if it isn't where I made my mistake?

Thank you very much
by the way, chocolatelover, writing out sentences to describe mathematical symbols and such makes your post tedious to read, at least for me. if you can't use LaTex, try these shorthands, it will make your posts more readable AND save you a lot of typing

instead of: the square root of x
type: sqrt(x)

instead of: the integral of xdx
type: int(x)dx

instead of: the integral from a to b of xdx
type: int_{a}^{b} (x)dx

OR

int (x)dx ...from a to b (keep the limits at the end of the line after you type everything else)

example:

instead of typing, as you did: integral 1 to the square root of 3 arctan(1/x)dx=xtan^-1(1-x)]one to the square root of 3-integral 1 to the square root of 3 x/1+x^2dx

you could type: int_{1}^{sqrt(3)}arctan(1/x)dx = xarctan(1/x) - int_{1}^{sqrt(3)}[x/(1+x^2)]dx

it is understandable that the limits for xarctan(1/x) is from 1 to sqrt(3), if you want to emphasis it, leave off the limits to the end

int [arctan(1/x)]dx = xarctan(1/x) - int [x/(1+x^2)]dx ...from 1 to sqrt(3)

now, see how nicer and shorter that is?

Or better yet, you could learn to use LaTex, it's not as hard as you may think. See our LaTex Tutorial

8. Originally Posted by chocolatelover

u=arctan(1/x)
du=[1/1+x^2]ln(x)
as red_dog and Krizalid said.

u = arctan(1/x)

=> du = -1/(1 + x^2)

why is that?

By the chain rule: $\frac {d}{dx} \arctan u = \frac {1}{1 + u^2}u'$

So, $\frac {d}{dx} \arctan \left( \frac {1}{x} \right) = \frac {1}{1 + \left( \frac {1}{x} \right)^2} \cdot \frac {-1}{x^2}$

............................ $= - \frac {1}{1 + x^2}$

9. Is this correct so far?

u=arctan(1/x)
du=-1/1+x^2
dv=dx
v=x

=tan^-1(x/)(-1/1+x^2)]1-sq. root 3-integr.1-squ.3-1/1+x^2dx
=tan^-1(1/sq.3)(-1/1+ sq. 3^2-tan^-1(1/1)(-1/1+1^2)-integ. 1-sq.3 -1/1+x^2dx
=-.131-.785-integ.1-sq.3-1/1+x^2dx

Thank you

10. Originally Posted by chocolatelover
Is this correct so far?

u=arctan(1/x)
du=-1/1+x^2
dv=dx
v=x

=tan^-1(x/)(-1/1+x^2)]1-sq. root 3-integr.1-squ.3-1/1+x^2dx
=tan^-1(1/sq.3)(-1/1+ sq. 3^2-tan^-1(1/1)(-1/1+1^2)-integ. 1-sq.3 -1/1+x^2dx
=-.131-.785-integ.1-sq.3-1/1+x^2dx

Thank you
no, that's not correct. and what you typed makes no sense, did you read through the last post i made?

Here's how to do it.

Using
$u = \arctan \left( \frac {1}{x} \right) \implies du = - \frac {1}{1 + x^2}$

$dv = dx \implies v = x$

We proceed using integration by parts

$\int_{1}^{\sqrt {3}} \arctan \left( \frac {1}{x} \right)~dx = \left. x \arctan \left( \frac {1}{x} \right) \right|_{1}^{\sqrt {3}} + \int_{1}^{\sqrt {3}} \frac {x}{1 + x^2}~dx$

for the last integral, we use the substitution: $u = 1 + x^2$

$\Rightarrow \int_{1}^{\sqrt {3}} \arctan \left( \frac {1}{x} \right)~dx = \left. x \arctan \left( \frac {1}{x} \right) + \frac {1}{2} \ln (1 + x^2) \right|_{1}^{\sqrt {3}}$

I leave plugging in the limits to you

11. Hi,

"arctan" is the same thing as tan^-1 right?

Can you please explain what you did? How did you get + integral? Isn't the formula [tex]\int[\math] a to b g'(x)dx=f(x)g(x)]a to b - [tex]\int[\math] a to be g(x)f'(x)dx

How did you get the x for =xarctan? Isn't that supposed to be f(x) wich is arctan(1/x)?

Where you said "we use the substitution u: 1+x^2" is du=x
dv=2x
v=x^2/2

To finish the problem do I need to plug in 1 and [tex]sqrt{x}[\math] to every x and subtract the two?

Thank you

12. again, you put the wrong slash on the [tex] tag! please follow instructions carefully when we type them, we don't do it just for fun, we do it to help you
Originally Posted by chocolatelover
Hi,

"arctan" is the same thing as tan^-1 right?
yes

Can you please explain what you did? How did you get + integral? Isn't the formula [tex]\int[\math] a to b g'(x)dx=f(x)g(x)]a to b - [tex]\int[\math] a to be g(x)f'(x)dx
the formula has a minus sign in it, yes, but du is negative, a negative times a negative is positive

How did you get the x for =xarctan? Isn't that supposed to be f(x) wich is arctan(1/x)?
The formula for integration by parts is:

$\int u~dv = uv - \int v~du$

$u = \arctan \left( \frac {1}{x} \right)$ and $v = x$

thus $uv = x \arctan \left( \frac {1}{x} \right)$

Where you said "we use the substitution u: 1+x^2" is du=x
dv=2x
v=x^2/2
no. integration by substitution is different from integration by parts, there is no need to find dv and v when doing substitution

To finish the problem do I need to plug in 1 and [tex]sqrt{x}[\math] to every x and subtract the two?

Thank you
you plug in $\sqrt {3}$ first and then minus the answer you get when you plug in 1

this is what the Fundamental Theorem of Calculus says. It's a very important theorem, please look it up in your text, you must know it

13. you plug in first and then minus the answer you get when you plug in 1
In other words, sqr. 3arctan(1/sqr3)+1/2ln(1+sqr. 3^2)-[1arctan(1/1)+1/2ln(1+1^2) right? My answer will be a decimal number, right? I won't have pie or anything like that, right?

Thank you

14. Originally Posted by chocolatelover
In other words, sqr. 3arctan(1/sqr3)+1/2ln(1+sqr. 3^2)-[1arctan(1/1)+1/2ln(1+1^2) right? My answer will be a decimal number, right? I won't have pie or anything like that, right?

Thank you
yes, it will be a decimal if you evaluate it. but you can leave the answer "exact" if you want. meaning no decimals. just work out the arctan's and simplify the ln's

15. just work out the arctan's and simplify the ln's
You can't simplify this any farther, can you?

Sqr. 3 acr tan(1/sqr. 3)- arctan(1)+1/2ln(4)-1/2ln(1)

If you can simplify it, can you show me how to?

Thank you

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