CALC 2 help integration?

Question

This problem has been blowing me mind for a soild hour can someone can please help me so i can sleep tonight

integrate:

5 / ((3e^x)-2) dx

Answer 1

Make the substitution u = 3exp(x), so that du = 3 exp(x) dx. This gives dx = 1/3 du/u, and so the integral becomes

(1/3) Integral du / u(u - 2).

This can be solved using partial fractions:

1 / u(u - 2) = (-1/2) / u + (1/2) / (u - 2),

so the integral becomes

(1/6) Integral [ 1/(u - 2) - 1/u ] du,

of which the antiderivative is

1/6 [log|u - 2| - log|u|] + c
= 1/6 log|(u - 2)/u| + c.

Now substitute u = 3 exp x:

= 1/6 log|3 (exp x) - 2 / 3 exp x| + c.

You might be able to simplify the answer a little more.

Answer 2

Let u = 3e^x - 2
Then du = 3e^x dx

u = 3e^x - 2 → 3e^x = u+2

So du = (u+2) dx
du/(u+2) = dx

and you get

integral of 5/[u(u+2)] du

Partial fractions time, right? I will leave that to you.

Note: your answer will have ln(3e^x) in it. This can be simplified:

ln(3e^x) = ln 3 + ln(e^x) = ln 3 + x

The ln 3 is just a constant which can be "absorbed" in the "+c" term.

You could make that x term "pop out" immediately by writing the original integrand as

(5/2)[3e^x - (3e^x - 2)]/(3e^x - 2) = (5/2)[3e^x/(3e^x - 2) - 1]

before using the u-substitution above. But I would never have thought to do that.

Answer 3

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