Question

definite integral (tan^-1 x) / (x^2 + 1). upper bound = infinity, lower bound = 1. use improper integral formula. show steps.

Answer 1

\[
u =\tan^{-1} x \\
du = \frac {dx}{1+x^2}\\

\int_{\frac \pi 4}^{\frac \pi 2} u du=\frac{3 \pi ^2}{32}

\]

Answer 2

how'd you get those bound values?

Answer 3

Initially the integral was: \[\int\limits_{x=1}^{x=\infty} ... \]
But then you changed to from x's to u's so we also had to change the bounds
Since \(u = \tan^{-1} x \) we just plug in each bound to that equation to get the new bounds of the integral.
\[\tan^{-1} \infty = {\pi \over 2}\]
\[\tan^{-1} 1 = {\pi \over 4}\]

Answer 4

hmm, yes, that much makes sense.

Answer 5

Do you need any more details?

Answer 6

no, that's good thanks