`int_0^(log2) e^(2x) dx`

Suppose that f,f^prime and f prime prime are continuous on [0, ln 2] and that f (0) = 0, f prime(0) = 3, f(In 2) = 6,f prime(ln 2) = 4 and int_0^(ln 2) e^(-2x)*f(x)dx=3. Find the value of int_0^(ln2) e^(-2x)*f prime prime(x)dx.

Let a=int_0^(log2) (2e^(3x)+e^(2x)-1)/(e^(3x)+e^(2x)-e^x+1)dx , then 4e^a =

What is int_(0)^(2) e^(ln) x dx equal to ?

The value of the integral int_0^(log5) (e^(x)sqrt(e^(x)-1))/(e^(x)+3)dx , is

int_(0)^(log 2)(e^(x))/(1+e^(x))dx=

" (2) "int e^(x log a)e^(x)dx=......+c

Given int_(1)^(2) e^(x^(2))dx=a , the value of int_(e )^(e^(4)) sqrt(log_(e )x)dx , is

Evaluate: int_0^1log(x^2+x+1)dx

int e^[2log e^(x)]dx