Solving 100 integrals stream day 7

[@CalculusIsFun1]

I’m editing this as I follow along.
for the first one, multiply by e^-x on top and bottom to force a trig sub. It becomes (e^-x)/(1 + e^-2x) let u = e^-x so du = -e^-xdx
the e^-x cancel and you are left with the integrals of -1/1 + u^2 = -tan^-1(u) =
-tan^-1(e^-x) + C
This can also be written as cot^-1(e^-x) + C (either will satisfy the integral).
EDIT: for the second one, whenever there’s an integral of a logarithm involved you should always think integration by parts.
Set u = log2(x) so du = 1/xln(2) set dv = 1dx so v = x
so it’s xLog2(x) - integrals of 1/ln(2) =
xLog2(x) - x/ln(2) + C