Asked by J
Integral of t^2/(t^2+b^2)^3/2 dt
Answers
Answered by
oobleck
use integration by parts, with
u = t
dv = t/(t^2+b^2)^3/2 dt
and everything drops out nicely, and you end up with
sinh<sup><sup>-1</sup></sup>(t/b) - t/√(t^2+b^2) + C
u = t
dv = t/(t^2+b^2)^3/2 dt
and everything drops out nicely, and you end up with
sinh<sup><sup>-1</sup></sup>(t/b) - t/√(t^2+b^2) + C
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