inteGRATE IT

use the property
f(a)=f(a+b-x)
u will get ur answer

one more pls.............
∫[x²-x+1] where [] denotes integer function from 0 to 2

pi*ln2/8

1+ tan(pi/4-x)= 2/(1+tanx)

I=₀∫^pi/4ln2dx - I

2I=pi/4*ln2

HINT for #3
look when the GINT beomes 1
and when it becomes 2
u have to substitute the values according to that

WARNING : DON'T BLINDLY PUT LIMITS AS 0 TO 1 AND 1TO2

ya abhi u r rite pls post me the soln

thanks

i got
wehere i commited mistake

i ve posted it now....see#5

1)

u will get

∫₀^pi/4ln(2/1+tanx) =I

∫₀^pi/4ln(1+tanx) =I

adding both u get

2I=∫₀^pi/4 ln2

I=pi*ln2/8

₀²âˆ«[x²-x+1]

x²-x+1=1
when x=0 and at x=1

x²-x+1=2
x=1/2+√5/2

x²-x+1=3
x=1/2+3/2=2

∫₀¹0dx +^1/2+√5/2₁∫2dx +_1/2+√5/2²âˆ«3dx

u get 2(1/2+√5/2-1)+3(2-1/2+√5/2)

evaluate it to get ur answer