Andrea O. answered • 04/15/15

Tutor

4.9
(196)
Tutors in Math from Elementary Math to Calc, Hablo Español

∫(1/(x+1))-(8x)

^{1/3}dxFor this problem you can work with it in parts because you subtracting, you can also do it with addition but not multiplying or dividing

∫(1/(x+1)) dx - ∫(8x)

^{1/3}dxLets solve the first part

∫(1/(x+1)) dx

If you don't know the derivative of ln(x) is 1/x so for this problem the anti-derivative is:

ln|x+1| +C

_{1}Now for the second part

- ∫(8x)

^{1/3}dx for this you can leave the negative outside the integral so you don't have to worry about itso the formula for this is:

∫ ax

^{n}dx = ax^{n+1}/ n +1So for our problem

a = 8

n = 1/3

n+1= (1/3) + (3/3) = 4/3

The anti-derivative is:

=(8x

^{4/3}) / (4/3) but here you are dividing by a fraction so you get=3•8x

^{4/3}/ 4=6x

^{4/3 }+ C_{2}Now we can put all the parts together and don't forget to bring back the negative sign

ln|x+1| +C

_{1}- 6x^{4/3}- C_{2}now C

_{1}- C_{2 }will just equal another constant so I will just call that Cln|x+1| - 6x

^{4/3}+ C