Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Integration C4

Intergrate lnx/x^3 dx

I did = intergrate x(x^-3) dx

Let ln x=u

x^-3 dv/dx

v= x^-2/-2

du/dx= 1/x

Now I can't do the last bit! Help please.

I'm not sure I follow what you are doing but this is easily done by parts, setting u=lnx and dv/dx = 1/x^3

EDIT: Seems you have done this, then

\int{u\frac{dv}{dx}} = uv - \int{v\frac{du}{dx}}

(lnx)/(x^3)

let u = lnx, du/dx = 1/x
v = -1/(2x^2), dv/dx = x^-3

Integral(u*dv/dx) = u*v - Integral(v*du/dx)

Therefore Integral ((lnx)/x^3) = -(lnx/2x^2) + Integral (1/(2x^3))

= -(1/2x^2)((lnx) + (1/2)) + C

I'd double check by differentiating, I'm prone to the odd numerical error, like factors of a half or whatever.

EDIT the first: Double checked, is right.

Integration by parts. u = lnx dv/dx = x^-3. Too late.

OP

miml

I'm not sure I follow what you are doing but this is easily done by parts, setting u=lnx and dv/dx = 1/x^3

That's what I have done.

So du/dx = 1/x

and v = x^-2 all over -2

*Integrate :wink:

:wink:

(without the r between e and g)

So you've let

u= \ln x \Rightarrow \dfrac{du}{dx} = \dfrac{1}{x}

and you've let

\dfrac{dv}{dx} = \dfrac{1}{x^3} \Rightarrow v = -\dfrac{1}{2x^2}

, so you're most of the way there; now you just need to apply the integration by parts rule:

Unparseable latex formula:

\displaystyle \int u\dfrac{dv}{dx}\, \mbox{d}x = uv - \int \dfrac{du}{dx}v\, \mbox{d}x

Subbing in your values, you get

Unparseable latex formula:

\displaystyle \int \dfrac{\ln x}{x^3}\, \mbox{d}x = -\dfrac{\ln x}{2x^2} - \int -\dfrac{1}{2x^3}\, \mbox{d}x

Then this is just a standard integral. It might help to write it as

Unparseable latex formula:

\displaystyle \int -\dfrac{1}{2}x^{-3}\, \mbox{d}x

if you're struggling.

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