Question

Any suggestions how to solve this integral?
Will post below

Answer 1

\[\int\limits_{}^{}\sqrt{1+e ^{2x}}dx\]

Answer 2

okay? then? du=e^2x(2)

Answer 3

Omg, I'm dumb. Sorry, I'm tired.

Answer 4

I think you make a trig substitution here:

e^x=tan(theta)

Answer 5

Oh nice, il try that thanks

Answer 6

\[put ~ \sqrt{1+e ^{2x}}=t,e ^{2x}=t^2-1,2e ^{2x}dx=2t~dt\]
\[dx=\frac{ tdt }{ e ^{2x}}=\frac{ tdt }{ t^2-1 }\]

Answer 7

\[I=\int\limits \frac{ t^2-1+1 }{t^2-1 }dt\]
i think now you cansolve after making partial fractions.

Answer 8

\[I=\int\limits dt+\int\limits \frac{ 1 }{\left(t+1 \right)\left( t-1 \right) }dt\]

Answer 9

okay thanks, and If I am doing this integral to solve the arclength of y=lnx from x=1 to x=2, would my new limits be 0 and ln(2)

Sorry original should have been \[\int\limits_{}^{}\sqrt{1+e ^{2y}}dy\]

Answer 10

\[t=\sqrt{1+e ^{2x}},when x=1,t=\sqrt{1+e^2},if x=2,t=\sqrt{1+e^4}\]

Answer 11

Right I mean, the problem asked for the arclength of \[y=\ln(x) \] from x=1 to x=2

Some suggested to try integrating in terms of y instead. Do you know what I mean?

Answer 12

i don't got you.
tell me original problem.

Answer 13

I will explain,

Find the arclength of \[y=\ln(x)\] between x=1 and x=2

\[L= \int\limits_{1}^{2}\sqrt{1+(dy/dx)^{2}} dx\]

\[L=\int\limits_{1}^{2} \sqrt{1+x^2}dx/\left| x \right| \]

The solution like this takes quite a while but it is equivalent to the arc length in terms of y instead, so the limits must be in terms of y instead as well as function to integrate as dy.

Answer 14

That problem is okay though, If needed I can just do it in terms of x. Another integral I am having trouble with is \[\int\limits_{}^{}\sqrt{1+4x ^{2}}dx\]

Answer 15

\[L=\int\limits_{1}^{2}\frac{ \sqrt{1+x^2} }{ x }dx=\int\limits_{1}^{2}\frac{ \sqrt{1+x^2}xdx }{x^2 }\]
\[put \sqrt{1+x^2}=t,x^2=t^2-1,2xdx=2tdt\]
\[L=\int\limits_{\sqrt{2}}^{\sqrt{5}}\frac{ t.tdt }{t^2-1 }\]

Answer 16

\[i=\int\limits \sqrt{1+4x^2}.1dx=integrate~ by~ parts\]