63 - Calculus Questions Answers

1+x²+x⁴ = (1+x+x²)(1-x+x²)

now solve

g(x) = 3-5sin(2x-4)

for getting reminder put x³= x so

x¹³⁵ + x¹²⁵ - x¹¹⁵ + x⁵ +1 will give

 x⁴⁵ + x⁴¹*x² - x³⁸*x +x*x² +1

x¹⁵ + x¹³*x²*x² - x¹²*x²*x +x*x² +1

 x¹⁷ + x³ +1

x⁵*x² + x + 1

x⁷ + x + 1

x²*x + x +1 

x³ + x +1 

x+x+1

2x+1

now put x = 10

let lnt = x so dt/t = dx so dt = tdx = e^xdx

now ∫(ln t)dt/(t-1) = ∫xe^xdx/(e^x-1) = ∫x(e^x-1+1)dx/(e^x-1)

now solve 

let  √tanx = t

so sec²xdx/2√tanx = dt

so (1+t⁴)dx/2t = dt

so dx = 2tdt/(1+t⁴)

now solve

2ƒ(x) − 3ƒ(1⁄x) = x²   (1)

so 2ƒ(1/x) − 3ƒ(x) = 1/x²  (2)

multiply (1) to 2 and (2) to 3, then add the two equations, you will get f(x) then calculate f(3)  

it will be 0, assume it by considering that ∞ is a big number 

for ex.   0*(1111111111111111111111111) = 0

f(x)= log₄log₃log₂x 

so 0 < log₃log₂x < ∞

so 1 < log₂x < ∞ 

so 2 < x < ∞