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 = \int (\frac {r^3}{4+r^2})^{0.5} dr

 

[MathewsMD]

Your integral expression appears corrupted.

Are you perhaps trying to find

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

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!