Expert: Socrates - 7/21/2007

Question
Hello, I would be really extremely grateful for some help on this problem. I have a huge
equation with big numbers which I need to antidifferentiate in order to continue
the problem. I will not put the numbers in but this is its form...(I tried to
write it out how it looks as I am not sure if I could write it out legiably
otherwise.)

               a
_____________________________
((b+c sqrt(d+ex))/f) +g

so it is 'a' divided by all the stuff on the bottom. And the (b+c sqrt (d+ex) is
all divided by the f. And then the g is added to the whole rest of the
denominator.If you could maybe put in your own values for the constants and show
me how I could antidifferentiate this in terms of 'x' I would be so grateful! It would allow me to attempt the rest of the problem.

Bye, Belle.


Answer
First , do some algebra to combine the constant terms . Next, make a change of variable to eliminate the radical. To finish, add and subtract the same quantity in the numerator to make it look like the denominator , the antiderivative should then be obvious.

Here is an example

Find S 9/[(3 + 2(4x+5)^1/2)/4 + 1] dx

factor out the 9

9 S 1/[(3 + 2(4x+5)^1/2)/4 + 1] dx

divide the terms in 3 + 2(4x+5)^1/2 by 4

9 S 1/[3/4 + (1/2)(4x+5)^1/2 + 1] dx

add 1 to 3/4

9 S 1/[ 7/4 + (1/2)(4x+5)^1/2 ] dx

multiply numerator and denominator by 4 to eliminate the fractions

9 S 4/[ 7+2(4x+5)^1/2 ] dx

factor out the 4

36 S 1/[ 7+2(4x+5)^1/2 ] dx


to get rid of the radical, make the substitution 4x+5 = u^2

so,

x = (1/4)u^2 - 5/4

dx/du = (1/2)u

dx = (1/2)u du


now, substitute u^2 for 4x+5 and (1/2)u du for dx in the integral

36 S (1/[ 7+2(u^2)^1/2 ]) (1/2)u du

36 S (1/[7+2u]) (1/2)u du

factor out the 1/2

18 S u/[7+2u] du

we want 2u in the numerator, so multiply and divide by 2

9 S 2u/[7+2u] du

now add and subtract 7 in the numerator

9 S (7+2u -7)/[7+2u] du


9 S  (7+2u)/[7+2u]  - 7/[7+2u] du

9 S 1 - 7/[7+2u] du

9u - 63 S 1/[7+2u] du

9u - (63/2) S 2/[7+2u] du

you must recognize the derivative of the natural log of a linear expression


9u - (63/2) ln[7+2u]

now , u^2 = 4x+5 , so u = (4x+5)^1/2

make this substitution to get the antiderivative back to the x variable

9(4x+5)^1/2 - (63/2) ln[7+2(4x+5)^1/2]

This is an antiderivative for the original function. You can check this answer by taking the derivative.

I hope this helps.