Question

Integration by Parts problem

Answer 1

\[\int\limits_{}^{}\tan^{-1}xdx\]

Answer 2

try
u = arctan x
dv = dx

Answer 3

\[\let u = \tan^{-1} and du = \frac{ -1 }{ x^2 +1 }dx\]

Answer 4

then dv = dx
and
v = x

Answer 5

so, \[\tan^{-1} \frac{ 1 }{ x } x - \int\limits_{}^{}xdx = \tan^{-1} \frac{ 1 }{ x } x - .5x^2 \]

Answer 6

* du should be positve
du 1/(x²+1)

Answer 7

why is du positive?

Answer 8

that's the derivative of arctan x.
\[\frac{ d }{ dx }\tan^{-1} x=\frac{ 1 }{ 1+x^2 }\]

Answer 9

yes but chain rule

Answer 10

we need to take the derivative of the inside function, which is 1/x

Answer 11

no. that's an inverse function, not an exponent

Answer 12

\[(1/x)\prime = -1/x^2\]

Answer 13

i think it's treated as an exponent

Answer 14

1/x = x^-1 = -1/2 * x^2

Answer 15

ohh I'm sorry!! i forgot to properly write the initial problem. It's supposed to say \[\int\limits_{}^{}\tan^{-1} (\frac{ 1 }{ x }) dx\]

Answer 16

i apologize. that's totally my fault

Answer 17

oh ok

Answer 18

okay, ahah, so sorry about that! so besides that little mishap, does everything else look right?

Answer 19

yeah it looks good

Answer 20

okay thank you!

Answer 21

yw

Answer 22

Just for future reference.

My previous answer was wrong, here's the correct, complete solution:
\[\int\limits_{}^{}\tan^{-1} (\frac{ 1 }{ x })dx = x \tan^{-1}(\frac{ 1 }{ x }) - \int\limits_{}^{}x*\frac{ 1 }{ x^2 +1 }dx \]

at this point, you have to do u-subsition in the integral.
so, let u = x^2 + 1, which will make .5du = xdx
we can bring the - and the .5 out front to give:
\[x \tan^{-1} (\frac{ 1 }{ x }) + \frac{ 1 }{ 2 }\int\limits_{}^{}\frac{ du }{ u }\] \[
=x \tan^{-1} (\frac{ 1 }{ x }) + \frac{ 1 }{ 2 }\ln|u| \]
\[= x \tan^{-1} (\frac{ 1 }{ x }) + \ln |x^2 + 1| + C\]