Integrate 1/x(2/3) - Solve for 3 Cube Root 3

[Roodles01]

knowing the standard form for integration by parts is
∫ f(x)g'(x) dx = f(x)g(x) - ∫f'(x)g(x) dx

I have what is an innocuous looking part of an equation which I can't solve.

the f(x) part in this case is;
ln(5x) which is easy enough i.e. 1/x

the second part 1/(x^(2/3)) is the bit I can't solve.

The standard I have for
1/x is ln(x)+c
& the standard I have for xⁿ is (1/(n+1))xⁿ⁺¹+c

But these don't solve this for me

I have checked on WolframAlpha & NumberEmpire & they give the same answer
3 cuberoot 3

I have tried just this bit by itself & go t nowhere. Could someone help with how I should get 3 cuberoot 3, please.

 

[Dick]

1/x^(2/3)=x^(-2/3). So put n=(-2/3) into x^(n+1)/(n+1).

 

[Roodles01]

I'm being thick here, but doesn't (n+1)/(n+1) = 1

So x¹ = x ?

Sorry.

 

[Dick]

I'm being thick here, but doesn't (n+1)/(n+1) = 1

So x¹ = x ?

Sorry.

Ok, I'll be a little clearer. I meant the formula you referred to [itex]\frac{x^{n+1}}{n+1}[/itex].

 

[HallsofIvy]

Yes, it is true that x¹= x! However, the anti-derivative of x¹ is x¹⁺¹/(1+1)= x²/2.

I am surprised that you are being asked to use "integration by parts" but do not know how to integrate xⁿ.

 

[Roodles01]

I shall try to be more numerically erudite in future!

I think that sometimes I have to ask stupid questions when I have come to the end of my tether & I can't see the wood for the trees.

Practice makes perfect & asking stupid questions should embarrass me into remembering it properly.

Prepare for more along the same lines in the future.