Question

Integration using tables of integrals... Integrate e^2x arctan(e^x) dx

Answer 1

I said u = e^x so du would be e^x as well but im not sure if this is the right substitution...

Answer 2

\[ \large \int e^{2x}\arctan(e^x)\,dx \]
right?

Answer 3

yes

Answer 4

do u know integration by parts?

Answer 5

yes... Int udv = uv -int vdu

Answer 6

ok

Answer 7

you have to use int by parts? it says using table tho...

Answer 8

i have nothing like that in my table.

Answer 9

what about \[\int\limits u \tan ^{-1} u du\]

Answer 10

if u have it. then u should use it.

Answer 11

i just got stumpd because if i let the u = e^x then i cant sub it in for the e^2x, you know?

Answer 12

i know. we could use this
\[ \large u=\arctan e^x \]
then
\[ \large \tan u=e^x \]

Answer 13

give me few minutes. i'll work this with pencil on paper

Answer 14

ok ill attempt as well

Answer 15

I have the answer, the textbook ...

Answer 16

It says you should get \[\frac{ (e^{2x} +1) }{ 2 } \arctan(e ^{x}) - \frac{ e^x }{ 2 } +c \]

Answer 17

this is what you get if you use that formula i had above..... now, in order top get this answer the u would have to be e^x

Answer 18

this is what i got.

Answer 19

let \(u=\arctan e^x\) so \(\tan u=e^x\). from this
we have that \(\tan^2u=(e^x)^2=e^{2x}\)
also \(\ln\tan u=x\) and from this
\[ \large dx=\frac{1}{\tan u}\sec^2u\,du \]
ok so far?

Answer 20

ok..

Answer 21

now
\[ \large \int e^{2x}\arctan e^x\,dx=
\int \tan^2u\cdot u\cdot\frac{\sec^2u}{\tan u}\,du \]
\[ \large =\int\tan u\cdot u\cdot\frac{1}{\cos^2u}\,du
=\int\frac{\sin u}{\cos u}\cdot u\cdot\frac{1}{\cos^2u}\,du \]
\[ \large =\int\frac{u\cdot\sin u}{\cos^3u},du \]
ok?

Answer 22

ok

Answer 23

now using integration by parts
\[ \large \begin{alignat*}{2}
A&=u &\qquad dB&=\frac{\sin u}{\cos^3u}\,du\\
Da&=du &\qquad B&=\frac{1}{2\cos^2u}
\end{alignat*} \]

Answer 24

\[ \large \int\frac{u\cdot\sin u}{\cos^3u}\,du=
\frac{u}{2\cos^2u}-\int\frac{1}{2\cos^2u}\,du \]
\[ \large =\frac{1}{2}\left[u\cdot\sec^2u-\int\sec^2u\,du\right] \]
\[ \large =\frac{1}{2}\left[u\cdot\sec^2u-\tan u\right] \]
OK?

Answer 25

ok...

Answer 26

now replacing \(u=\arctan e^x\) we have finally
\[ \large \int e^{2x}\arctan e^x\,dx=\frac{1}{2}
\left[\arctan e^x\cdot\sec^2(\arctan e^x)-
\tan(\arctan e^x)\right] \]

Answer 27

but
\[ \large \tan(\arctan e^x)=e^x \]
and using \(\sec^2z=\tan^2z+1\) we have
\[ \large \sec^2(\arctan e^x)=\tan^2(\arctan e^x)+1=(e^x)^2+1=
e^{2x}+1 \]

Answer 28

therefore
\[ \large \int e^{2x}\arctan e^x\,dx=1/2[(e^{2x}+1)\cdot\arctan e^x-e^x] \]

Answer 29

ahhhhhh

Answer 30

it is weird that they asked u to do this "by tables".

Answer 31

see u some other time.

Answer 32

thank you very much

Answer 33

u r welcome