Evaluate using any method: Essentials Of Calculus

[Nawz]

Homework Statement

$$\int\frac{ [(lnx)^3 - 4(lnx)^2 + 5]}{x}$$dx

Homework Equations

The Attempt at a Solution

xlnx^4 - (4/3x)lnx^3 +5x^2

my attempt at a soultion was all wrong. The divided by x is confusing me on this one.

The right answer is: ((lnx)^4/4) -(4/3)(lnx)^3+5lnx+C

 

[rock.freak667]

(You forgot the 'dx' in your integral)


but if you let u=lnx, what is du =?

 

[Nawz]

It would be 1/x

What is the point of the dx after it too: is it just there for the added C?

I got this right now and it is close:

$$\int\frac{(u)^3-4(u)^2+5}{x}$$dx

$$\frac{(u)^4/4-(4/3)u^3+5x}{x}$$ + C

I guess that 5x is suppose to be a U but I don't know why. I still have it divided by x too?

 

[rock.freak667]

It would be 1/x

What is the point of the dx after it too: is it just there for the added C?

In your integral, it means what you are integrating with respect to.

so if du/dx = 1/x, then that means that du = (1/x)dx.

I got this right now and it is close:

$$\int\frac{(u)^3-4(u)^2+5}{x}$$dx

Do you see where you will have '(1/x)dx' to replace with 'du'?