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

Question

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

anonymous · Answer

$$
u =\tan^{-1} x \
du = \frac {dx}{1+x^2}\

\int_{\frac \pi 4}^{\frac \pi 2} u du=\frac{3 \pi ^2}{32}

$$

anonymous · Answer

how'd you get those bound values?

eyust707 · Answer

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.
$$\tan^{-1} \infty = {\pi \over 2}$$
$$\tan^{-1} 1 = {\pi \over 4}$$

anonymous · Answer

hmm, yes, that much makes sense.

anonymous · Answer

Do you need any more details?

anonymous · Answer

no, that's good thanks