calculus problem?

Question

for some reason i can't find a way to do it
find the integral of x/(x+4)

Answer 1

Divide first:

x/(x+4)=1 +[-4/(x+4)]

Then int(x/(x+4))=int(1)+(-4) *int(1/(x+4)=x- 4ln(|x+4|)+C

Always divide first when degree of the top is >= degree of the bottom

Answer 2

So you want to solve

Integral (x/(x + 4))dx

Rule of thumb: Whenever you're dividing polynomials, and the degree of the numerator is greater than or equal to the degree of the denominator, use long division. In this case, the degree of the numerator is 1 (1 is the highest power of x), and the degree of the denominator is 1.

We use long division, i.e

x + 4 (INTO) x + 0

I won't show you the details of how the long division is done, but you should end up with

x/(x+4) = -4/(x+4) + 1

Now we can integrate this:

Integral (-4/(x+4) + 1)dx ... splitting this up into two integrals, we get

Integral (-4/(x+4))dx + Integral (1)dx

Pulling out the constant -4, we get

-4 * Integral (1/(x+4))dx + Integral (1)dx

And now this becomes easy to solve.

-4 ln |x + 4| + x + C

Answer 3

write x/(x+4) = (x+4-4)/(x+4) = 1 - 4/(x+4). The integral should be obvious from here on.

Of course doing a u-substitution u=x+4 would also be fine.

Answer 4

Try substitution:

Let u = x+4 then x= u-4
du = dx

So anti (u-4) / u du
= anti (u/u) - (4/u) du
= anti 1 - 4/u du
= u - 4 * ln |u| + C
= (x+4) - 4 * ln|x+4| +C (**)

Check:

Derivative of (**) is 1 - 4 * 1/(x+4) = (x+4)/(x+4) - 4/(x+4) =x/(x+4)

Answer 5

Do the division of x/(x+4) first.

Then do the integration term by term.

Answer 6

x/(x + 4) = 1 - 4/(x + 4)

∫[x/(x + 4)]dx = ∫[1 - 4/(x + 4)]dx = x - 4 ln (x + 4) + C

Answer 7

x-4ln(Ix-4I)

x-4 is just x
and lnx = 1/x