Question

calculas.........help pll .............
URGENT...........

Answer 1

@amistre64 @reemii

Answer 2

What is the part at the start there? is that a 2I? Like this?\[\Large 2I\quad=\quad \int\limits_0^{\pi/4}\ln\left[1+\tan x +\frac{2}{1+\tan x}\right]dx\]

Answer 3

ya

Answer 4

So we can start by getting a common denominator in the brackets.\[\large 2I= \int\limits_0^{\pi/4} \ln\left[\frac{(1+\tan x)+\tan x(1+\tan x)+2(1+\tan x)}{1+\tan x}\right]dx\]

Answer 5

The top multiplies out to give us,\[\large 2I= \int\limits\limits_0^{\pi/4} \ln\left[\frac{\tan^2x+4\tan x+ 3}{1+\tan x}\right]dx\]
Which can be factored, Yay!\[\large 2I= \int\limits\limits\limits_0^{\pi/4} \ln\left[\frac{(1+\tan x)(3+\tan x)}{1+\tan x}\right]dx\]

Answer 6

Confused on anything up to that point? :o

Answer 7

i got till there ...
then