Integration - Can't find my mistake

[MathewsMD]

##\int arctan(\frac {1}{x})dx = I## when ##u = arctan(\frac {1}{x}), du = -\frac {1}{1+x^2} dx, dv = dx, v = x## and I got ##du## by chain rule since ##\frac {1}{1 + \frac {1}{x^2}} (-\frac {1}{x^2})## simplifies to ##-\frac {1}{1+x^2} dx## right?

##I = xarctan(\frac {1}{x}) + \int \frac {x}{1+x^2}dx##
##I = xarctan(\frac {1}{x}) + \int \frac {1}{x}dx + \int xdx##
##I = xarctan(\frac {1}{x}) + ln lxl + 0.5x^2 + C ##

This is not the correct answer, though, and I can't seem to find my exact error.

Any help would be great!

P.S. I have seen the identities to make arctan(1/x) = arctan(x) and using substation earlier, but am trying to use this method.

Second integration problem:

##I = \int \frac {r^3}{4+r^2}^0.5 dr## and if ##4 +r^2 = u## or ##r^2 = u - 4## then ##dr = \frac {1}{2r} du## so
##I = 0.5 \int \frac {u-4}{u}^0.5 du = 0.5 [\int u^0.5 - 4\int {1}{u}]##
##I = \frac {1}{3} u^{3/2}- 8u^{1/2}##

This is also wrong but I'm not quite sure why.

 

[LCKurtz]

##\int arctan(\frac {1}{x})dx = I## when ##u = arctan(\frac {1}{x}), du = -\frac {1}{1+x^2} dx, dv = dx, v = x## and I got ##du## by chain rule since ##\frac {1}{1 + \frac {1}{x^2}} (-\frac {1}{x^2})## simplifies to ##-\frac {1}{1+x^2} dx## right?

##I = xarctan(\frac {1}{x}) + \int \frac {x}{1+x^2}dx##
##I = xarctan(\frac {1}{x}) + \int \frac {1}{x}dx + \int xdx##

##\frac x {1+x^2} \ne x + \frac 1 x##.

 

[MathewsMD]

##\frac x {1+x^2} \ne x + \frac 1 x##.

General case: ##\frac {d}{dy} arctan(y) = \frac {1}{1 +y^2}## right?
So replacing y with 1/x
##= \frac {1}{1 + (1/x)^2}##
##= \frac {1}{\frac {x^2 + 1}{x^2}}##
##= \frac {x^2}{1+x^2}## [1]

Then applying chain rule to ##\frac {1}{x}##
##\frac{d}{dx}\frac {1}{x} = -\frac {1}{x^2}## [2]

So I multiply [1] by [2] to get

## -\frac {1}{1 + x^2}##

Oh, never mind, just realized you stated something completely different...

 

[MathewsMD]

##\frac x {1+x^2} \ne x + \frac 1 x##.

Just a question: If this was a definite integral and did not included 0 between the upper and lower limits, would it be allowed or still no? Since you have to find the general antiderivative either way, I guess it doesn't make sense...

Just wondering, what exactly would be the next step to take here? I seem to have made the same mistake in both questions.

 

[LCKurtz]

Just wondering, what exactly would be the next step to take here? I seem to have made the same mistake in both questions.

You had$$
x\arctan(\frac 1 x) +\int\frac x {x^2+1}~dx$$Do a ##u## substitution on the second integral.

 

[SteamKing]

second integration problem:

##i = \int \frac {r^3}{4+r^2}^0.5 dr## and if ##4 +r^2 = u## or ##r^2 = u - 4## then ##dr = \frac {1}{2r} du## so
##i = 0.5 \int \frac {u-4}{u}^0.5 du = 0.5 [\int u^0.5 - 4\int {1}{u}]##
##i = \frac {1}{3} u^{3/2}- 8u^{1/2}##

this is also wrong but I'm not quite sure why.

Your integral expression appears corrupted.

Are you perhaps trying to find

I = [itex]\int (\frac {r^3}{4+r^2})^{0.5} dr[/itex]

 

[MathewsMD]

Your integral expression appears corrupted.

Are you perhaps trying to find

I = [itex]\int (\frac {r^3}{4+r^2})^{0.5} dr[/itex]

Sorry, yes. I was in a hurry and forgot to fully check my script. My problem has been solved by the above answer. Thank you for all the help!