find the integral of e^x/[(e^x-3)(e^x+4)] dx
To find the integral of the given function, we can start by using partial fraction decomposition.
First, let's factorize the denominator:
(e^x - 3)(e^x + 4)
Next, we need to find the values of constants A and B that satisfy the equation:
A/(e^x - 3) + B/(e^x + 4) = e^x/[(e^x - 3)(e^x + 4)]
Multiplying through by the common denominator, we have:
A(e^x + 4) + B(e^x - 3) = e^x
Expanding and collecting like terms, we get:
(A + B)e^x + 4A - 3B = e^x
To match the coefficients of e^x, we have the following conditions:
A + B = 1 (coefficient of e^x on the right-hand side)
4A - 3B = 0 (constant term on both sides)
Solving these equations simultaneously, we find A = 3/7 and B = 4/7.
Now, we can rewrite the integral using the partial fraction decomposition:
∫(3/7)/(e^x - 3) dx + ∫(4/7)/(e^x + 4) dx
To integrate the fractions, we can use the natural logarithm function. Recall that:
∫1/(u) du = ln|u| + C
Applying this to our fractions, we get:
(3/7) ln|e^x - 3| + (4/7) ln|e^x + 4| + C
So, the integral of e^x/[(e^x-3)(e^x+4)] dx is:
(3/7) ln|e^x - 3| + (4/7) ln|e^x + 4| + C, where C is the constant of integration.