Make a substitution to express the integrand as a rational function and then evaluate the integral.?

Question

The integral of e^(2x) / [e^(2x) + 3e^x + 2] dx.

The answer should be:
ln[(e^x + 2)^2 / (e^x + 1)] + C .
Please show how you arrived at your final answer. Thanks for your help:)

Answer 1

Your substitution should be u = e^x.
Keep in mind that e^(2x) is the same as (e^x)(e^x) = (e^x)^2

If u = e^x, then
du = e^x dx

Integral (e^(2x) / [e^(2x) + 3e^x + 2]) dx

Changing it around a bit, to make something obvious:

Integral ( [e^x] / [e^(2x) + 3e^x + 2] )(e^x dx) )

What I did: I separated e^(2x) into e^x and e^x, putting one close to the dx, because du = e^x dx.

Now, we replace.

Integral (u / (u^2 + 3u + 2)) du, or

Integral (u / [(u+1)(u+2)]) du

Here, we use partial fractions, noting that

u / [(u+1)(u+2)] = A/(u+1) + B/(u+2).

We can either solve A and B now or later. I want to solve them later. As a result, our integral becomes

Integral (A/(u+1) + B/(u+2))du

Which we can split up into two integrals.

Integral (A/(u+1)) du + Integral (B/(u+2)) du

AND, we can pull out A and B, because they're constants.

A * Integral (1/(u+1))du + B * Integral (1/(u+2))du

These integrals are easy to solve now.

A [ln|u+1|] + B[ln|u+2|] + C

Our next goal is to solve for A and B. We do this with this statement:

u / [(u+1)(u+2)] = A/(u+1) + B/(u+2)

If we multiply both sides by (u+1)(u+2), the denominators all vanish, leaving behind factors to multiply.

u = A(u+2) + B(u+1)

Remember that the above statement is true ***FOR ALL u***. That means we can substitute whatever we like in the above equation and still solve for A and B.

Let u = -2:

-2 = A(-2 + 2) + B(-2 + 1)
-2 = A(0) + B(-1)
-2 = -B
B = 2

Let u = -1

-1 = A(-1 + 2) + B(-1 + 1)
-1 = A(1) + B(0)
A = -1

Therefore, A = -1 and B = 2. Make the appropriate substitutions for here:

A * [ln|u+1|] + B[ln|u+2|] + C
(-1)[ln|u+1|] + 2 [ln|u+2|] + C

BUT, at the start, we let u = e^x, so our final answer is

(-1) [ln|e^x + 1|] + 2 [ln |e^x + 2|] + C

We can arrange this and use log properties to combine into a single log. Let's do that.

Rearranging the first two terms, and dissolving the (-1) into a negative symbol, we get:

2[ ln|e^x + 2| ] - [ln|e^x+1|] + C

Using the property: b*ln(a) = ln(a^b)

ln ( [e^x + 2]^2 ) - ln ( |e^x + 1| ) + C

Using the property: ln(a) - ln(b) = ln(a/b)

ln ( [e^x + 2]^2 / |e^x+1|) + C

Answer 2

Let u=e^x, then du=e^x dx. The integrand becomes u/(u^2 +3u+2) du.

Now factor u^2+3u+2=(u+2)(u+1) and do partial fractions to find that u/[(u+2)(u+1)]=-1/(u+1) +2/(u+2). This gives an integral of

ln[(u+2)^2/(u+1)] +C.
Now let u=e^x to get the answer.