compute the antiderivative

Question

antiderivative ((1/x+1)-(8x)^(1/3)) dx

 

Please show all the steps!!

Answer 1

∫(1/(x+1))-(8x)^1/3 dx

 

For 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 dx

 

Lets 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| +C₁

 

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 it

 

so the formula for this is:

 

∫ axⁿ dx = axⁿ⁺¹ / n +1

 

So 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 + C₂

 

Now we can put all the parts together and don't forget to bring back the negative sign

 

ln|x+1| +C₁ - 6x^4/3 - C₂

 

now C₁ - C₂ will just equal another constant so I will just call that C

 

ln|x+1| - 6x^4/3 + C