Help with these calc problems: taking the anti-derivative / integral.....?

Question

4e^x / 5+e^x dx

I need to find the integral (or anti-derivative) of the problem.
please help

and

the integral of (x+3) / (x^2 + 6x + 8)

in the second problem that you solved, what did u do with the (x+3)? explain how u got the answer please.

and the first question is:

(4e^x) / (5 + e^x).....find integral

Answer 1

The second is easy. The derivative of x^2+6x+8 is 2x+6, so rewrite it as (1/2)((2x+6)/(x^2+6x+8)) and it's in the form of ln(x) dx: (1/2)ln(x^2+6x+8)
On the first one, you need to add some parentheses; maybe you're asking about (4e^x/5)+e^x, or maybe (4e^x)/(5+e^x).

Answer 2

You use substitution for these.. You usually have one bit that's complicated, but can be substituted with something simpler that can be integrated easily. Your derivative (dx) also changes. Here, take a look:

First one: (d is derivative and int means integral, okay?)
Now, as you can see, it's the denominator that's making things complicated. So let's try substituting it with something simpler.

let u=5+e^x

Taking the derivative of this substituted equation
=> du=e^x dx

Which is part of the numerator of your question

So you can write the question like this:
=> int [4e^x /( 5+e^x) dx] = int [4/u du]
= 4 ln (u) + constant
= 4 ln (5+e^x) + constant
Which is your final answer. Don't forget to resubstitute your assumption!
_______________________________
Second one:

let u = (x^2 + 6x + 8)
=> du= (2x+6) dx
=> du/2 = (x+3) dx
Which is part of the numerator of your question, again.

So you can write that as...
=> Int[du/(2u)]
=0.5 ln (u)+ const.
=0.5 ln (x^2 + 6x + 8) + const.

That's all! Hope it helps.

Answer 3

Question 1
I = 4 ∫ e^x / (5 + e^x).dx
I = 4.ln(5 + e^x) + C

Question 2
(x + 3) / (x + 4).(x + 2)
(x + 3) = A / (x + 4) + B / (x + 2)
(x + 3) = A.(x + 2) + B.(x + 4)
1 = A + B
3 = 2A + 4B
2 = 2A + 2B
1 = 2B
B = 1/2
A = 1/2
I = ∫(x + 3) / [(x + 4).(x + 2)] .dx
I = (1/2).∫ 1 / (x + 2).dx + (1/2).∫1 / (x + 4).dx
I = (1/2).log (x + 2) + (1/2).log (x + 4) + C
I = log(x + 2)^(1/2) + log(x + 4)^(1/2) + log K
I = log [K [(x + 2).(x + 4)]^(1/2) ]