Indefinite integral: ∫ (4+u^2)/u^3 du?

Question

All I could think is to change it to ∫ (4/u^3)+u^(-1) du
I don't know what to do with the (4/u^3) part.
The other problem would be 4^0/0

I think I'm suposed to use natural log maybe... somehow? help! :-)

Answer 1

You are absolutely correct in changing it into two fractions.

∫ ( (4 + u^2)/u^3 du )

∫ ( (4/u^3 + 1/u) du )

Now, we split this into two integrals.

∫ (4/u^3 du) + ∫(1/u) du

Factor out the 4.

4 ∫ (1/u^3) du + ∫(1/u) du

Change the form of 1/u^3.

4 ∫(u^(-3) du ) + ∫ ( (1/u) du )

Use the reverse power rule on the first integral, and directly solve the second integral.
Recall that the reverse power rule goes as follows:
The integral of x^n, for n not equal to -1, is (1/[n + 1])x^(n + 1).
If n happens to be equal to -1 (which is true, in the second integral), the integral is just the natural log.

4 (-1/2)u^(-2) + ln|u| + C

Which simplifies as

-2 u^(-2) + ln|u| + C

-2/u^2 + ln|u| + C

Answer 2

∫ (4+u^2)/u^3 du
= ∫ 4/u^3 + 1/u du
= -2/u^2 + ln|u| + c

Answer 3

integral (4/u^3) du = integral (4u^-3) du = 4u^-2/(-2) + c = -2/u^2 + c

integral 1/u du = ln u + c

Now, add both integral and get:

∫ (4+u^2)/u^3 du = -2/u^2 + ln |u| + c

Answer 4

∫ (4+u^2)/u^3 du= ∫ (4u^3+u^5)/ du?=u^4 +u^6 /6. answer