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Fundamental Theorem of Calculus, Will Rate

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Question:

Answers:

pretty4u

Novice

Karma Points: 44

Date Posted: 7/23/2008 10:49:54 PM Status: Live

Asker's Rating: Helpful

Response:

Use integration by parts

u=arctan(t) du=

dv=dt v=t

=tarctan(t)-

Use u-substitution

u=1+

du=2tdt

du=tdt

=tarctan(t)-

ln(1+

)

=[1/x(arctan(1/x))-1/2ln{1+(1/x)^2}]-[2arctan(2)-1/2ln{1+(2)^2}]

=0-[2arctan(2)-1/2ln{1+(2)^2}]

=1/2ln(5)-2arctan(2)

Achilles

Sage

Karma Points: 5,455

(Melbourne University)

Date Posted: 7/24/2008 10:10:54 AM Status: Live

Asker's Rating: Helpful

Response:

Let u=arctan(t)....................................................................>     du=dt=tarctan(t)-
dv=dt.....................................................................................>    v=t

Integration by parts is uv-
∴=t.arctan(t)-

                      =tarctan(t)-ln(1+t²)

          =[1/x(arctan(1/x))-1/2ln(1+(1/x)²)]-[2arctan(2)-1/2ln(1+(2)²)]

           =0-[2arctan(2)-1/2ln(1+(2²)]

           =1/2ln(5)-2arctan(2)