What is the improper integrals 1/((8x^2)+(1)) from 0 to infinity ?

1 Answer
May 5, 2018

018x2+1dx=π42

Explanation:

To start with, we slightly rewrite the integral:

018x2+1dx=01(8x)2+1dx.

Now, we may view 8x as an inner function, which is why this is the same as

1801y2+1dy=

=18[tan1y]0=

=18[π20]=

=π28=

=π42.