Evaluate the following Integral (Calculus 2)?

Question

Hey all I'm having some trouble with evaluating this integral. Any help is deeply appreciated.

Evaluate the following integral:
int=integral

int[1/(5(e^x)+2) dx]= ?

Thanks in advance

Answer 1

First, we perform a substitution:
u=5e^x+2, du=5e^x dx, dx=du/(u-2)
So our integral becomes:
∫1/(u(u-2)) du
Now this is the perfect form for partial fraction decomposition. We solve:

A/u + B/(u-2) = 1/(u(u-2))
A(u-2) + Bu = 1
(A+B)u - 2A = 1
-2A = 1, A+B=0
A=-1/2, B=1/2

So our integral becomes:

∫-1/2/u + 1/2/(u-2) du
1/2∫1/(u-2) - 1/u du
1/2(ln (u-2) - ln u) + C

Substituting our original value for u:

1/2(ln (5e^x) - ln (5e^x + 2)) + C

And simplifying:

ln (5e^x/(5e^x+2))/2 + C

Answer 2

First, split it up into two separate integrals:

Integral(1/5e^x)dx + Integral(2)dx

Pull out the constants to put outside of the integrals:

1/5 * Integral (1/e^x)dx + 2 * Integral(1)dx

Note that the reciprocal of an exponent becomes a negative exponent, therefore 1/e^x = e^(-x). The right integral is easy to solve. Thus, we get.

1/5 * Integral (e^(-x))dx + 2x

The integral of e^x is e^x, and you'll learn through simple substitution and differentiation verification that the integral of e^(-x) is - e^(-x). Thus, our answer is

1/5 * (- e^(-x) ) + 2x + C

or

-1/5 * e^(-x) + 2x + C

Answer 3

1/ int of (u^5)du = u^6 / 6 = (1+sint)^6 / 6 + c 2/ g ' (x) = (2x+2) cos(x^2+2x)

Answer 4

ln(arctan1/5 e^x)+2