Thread: Integral of x.arctan(x) with hint with hint

1. Integral of x.arctan(x) with hint with hint

Hi, I have the problem to find:

integral of - x.arctan (x)

the hint / first step is pointed to by first asking to find:

integral of - x^2 / ( 1 + x^2 )

I have found the answer to be:

1/2 ( x^2 +1 ) . arctan (x) - 1/2.x

I cannot deduce the working to get here though. Any help much appreciated... thanks

2. We have

$\displaystyle \int x \tan^{-1}(x)~dx$

$\displaystyle = \frac{1}{2} \int \tan^{-1}(x) ~d(x^2+1)$

We now use integration by parts ,

$\displaystyle u = x^2 + 1 ~,~ v = \tan^{-1}(x)$

$\displaystyle = \frac{1}{2} \int v ~du$

$\displaystyle = \frac{1}{2} [ (u)v - \int u dv ]$

$\displaystyle = \frac{1}{2} [ (x^2 + 1 ) \tan^{-1}(x) - \int (x^2 + 1 )d( \tan^{-1}(x)) ]$

$\displaystyle = \frac{1}{2} [ (x^2 + 1 ) \tan^{-1}(x) - \int (x^2 + 1 ) \frac{dx}{x^2+1}]$

$\displaystyle = \frac{1}{2} [ (x^2 + 1 ) \tan^{-1}(x) - \int dx ]$

$\displaystyle = \frac{1}{2} [ (x^2 + 1 ) \tan^{-1}(x) - x ] + C$

3. I'm sorry, your solution is very good and most succinct but to gain full marks for my assignment think i have to use the integral of
x^2 / ( 1+x^2 )

I have, from more directly integrating by parts of x.arctan (x) where

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

dv=x giving v=1/2.x^2

in formula int u.dv = u.v. - int v.du

shows the significance of the x^2 / ( 1+x^2 ) term by -

1/2.x^2.arctan(x) - 1/2. int x^2 / ( 1+x^2 )

I am glad to have seen your method, I have not seen that technique before (your very first step of sub know integral of x^2 + 1, does it have a name?) I have tried to see how it might be relavent to my current position as I have attempted to explain - but failed again unfortunately. If you could again assist i would be most humbled. Also, are you inserting Images made in / with word equation editor?

many thanks !

4. Originally Posted by Hytholoday71 I'm sorry, your solution is very good and most succinct but to gain full marks for my assignment think i have to use the integral of
x^2 / ( 1+x^2 )

I have, from more directly integrating by parts of x.arctan (x) where

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

dv=x giving v=1/2.x^2

in formula int u.dv = u.v. - int v.du

shows the significance of the x^2 / ( 1+x^2 ) term by -

1/2.x^2.arctan(x) - 1/2. int x^2 / ( 1+x^2 )

I am glad to have seen your method, I have not seen that technique before (your very first step of sub know integral of x^2 + 1, does it have a name?) I have tried to see how it might be relavent to my current position as I have attempted to explain - but failed again unfortunately. If you could again assist i would be most humbled. Also, are you inserting Images made in / with word equation editor?

many thanks !
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