Solving an integral: int [arct(e^x) / e^(2x)] dx

[Danilo432]

HELP

 

[stapel]

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HELP

What are your thoughts? What have you tried? Where are you stuck?

Please be complete. Thank you!

Eliz.

 

[BigBeachBanana]

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HELP

Try integration by parts.

 

[skeeter]

let [imath]t = e^x \implies dt = e^x \, dx \implies dx = \dfrac{dt}{t}[/imath]

[imath]\displaystyle \int \dfrac{\arctan{t}}{t^3} \, dt[/imath]

[imath]u = \arctan{t} \implies du = \dfrac{dt}{1+t^2}[/imath]

[imath]dv = \dfrac{dt}{t^3} \implies v = -\dfrac{1}{2t^2}[/imath]

[imath]\displaystyle \int u \, dv = u \cdot v - \int v \, du[/imath]

proceed with the above integration by parts as recommended by the Banana. Note that another iteration of integration by parts for the (v du) integral may be needed.

 

[Danilo432]

let [imath]t = e^x \implies dt = e^x \, dx \implies dx = \dfrac{dt}{t}[/imath]

[imath]\displaystyle \int \dfrac{\arctan{t}}{t^3} \, dt[/imath]

[imath]u = \arctan{t} \implies du = \dfrac{dt}{1+t^2}[/imath]

[imath]dv = \dfrac{dt}{t^3} \implies v = -\dfrac{1}{2t^2}[/imath]

[imath]\displaystyle \int u \, dv = u \cdot v - \int v \, du[/imath]

proceed with the above integration by parts as recommended by the Banana. Note that another iteration of integration by parts for the (v du) integral may be needed.

Thank you so much. Is there a way to mark your answer as a solution? I am new on this website..

 

[mmm4444bot]

Is there a way to mark your answer as a solution?

Hello. That option is available, but only when it has been set up at the time the thread was created. Please see post#2 in this thread about forum features. ?
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