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

Question

The integral (with upper limit 16 and lower limit 9) of (sqrt. x) / (x-4) dx.
The correct answer should be 2 + ln(25/9) .
Please show how you arrived at your answer. Thanks so much for your help!

Answer 1

We first make the following substitutions:
(1) u = sqrt(x)
(2) by squaring both sides of (1), u^2 = x
(3) by differentiating (2), 2*u*du = dx.
(4) since we are doing a full substitution, we need to replace the upper and lower limits. Thus, to replace the upper limit, we have u = sqrt(16) = 4 and the lower limit becomes u = sqrt(9) = 3.

The integral is now:
2*u*u*/(u^2 - 4) du with upper limit 4 and lower limit 3
Simplifying we have:
2u^2/(u^2 - 4) du with the same limits.

We now notice that 2u^2/(u^2 - 4) = 2 + 8/(u^2 - 4) by using polynomial division. I.e., u^2 - 4 divides into 2u^2 giving a quotient of 2 and a remainder of 8. (I would like to show this, but it's hard to show polynomial division on this.)

Thus, our integral is now:
(2 + 8/(u^2 - 4)) du with the same limits as above.

We now use partial fraction decomposition on 8/(u^2 - 4) to obtain
2/(u-2) - 2/(u+2) (see below). Our integral is now, with the same limits of lower limit of 3, upper limit of 4
2 + 2/(u-2) - 2/(u+2) du.

Integrating term by term, we have:
2u + 2ln(u-2) - 2ln(u+2) evaluated from 3 to 4. Thus we have:
(2(4) + 2ln(4-2) - 2ln(4+2)) - (2(3) + 2ln(3-2) - 2ln(3+2)) =
(8 + 2ln(2) - 2ln(6)) - (6 + 2ln(1) - 2ln(5)) =
8 + 2ln(2) - 2ln(6) - 6 - 2ln(1) + 2ln(5) =
8 + 2ln(2) - 2ln(6) - 6 - 0 + 2ln(5) =
2 + ln(4) - ln(36) + ln(25) =
2 + ln(4) + ln(25) - ln(36) =
2 + ln(100) - ln(36) =
2 + ln(100/36) =
2 + ln(25/9).

Recall:
(1) ln(M) + ln(N) = ln(M*N) and ln(M) - ln(N) = ln(M/N)
(2) ln(1) = 0
(3) the partial fraction decomposition of 8/(u^2 -4):
8/(u^2 - 4) = 8/(u+2)(u-2) = A/(u+2) + B/(u-2).
Multiplying both sides of the above equation by (u+2)(u-2), we get
8 = A(u-2) + B(u+2).
Setting u = 2 gives us
8 = B*4 --> B = 2
Setting u = -2 give us
8 = A*(-4) --> A = -2.
Thus 8/(u^2 - 4) = -2/(u+2) + 2/(u-2)

Thanks.

Answer 2

Let u = sqrt(x). Then
u^2 = x,
u^2 - 4 = x - 4, therefore
2u du = dx

Your upper and lower bound get replaced by the substition.
When x = 16, u = 4
When x = 9, u = 3.
New bounds of integration: 3 to 4.

Integral ( [u / (u^2 - 4)] 2u du )
2 * Integral ( [u^2/(u^2 - 4)] du ), from 3 to 4.

Now, it's expressed as a rational integral.
I'll leave the second half up to you, because it involves trigonometric substitution (u = 2sec(theta)) and that takes long.

Answer 3

i got it in 3 [e-mail], and i will send u the scaned paper, k?