Intergration help! (1 Viewer)

iishorty · Jul 2, 2012

I'm currently going through the NSGHS 2009 mathematics 2 unit trial and got stuck at question 9d,ii. I've looked at the solutions and don't understand the working out for the question. It involves find the integral of lnx/x^2 from e to 1, using the result in the previous part which was (1-lnx)/x^2.

The solution says that Inx/x^2 = -[(1-lnx)/x^2 - 1/x^2] and uses this to carry out the integration, but I don't understand how that equals that.

Can someone please help me understand it or give me another way of doing the question!

Thanks

rolpsy · Jul 2, 2012

iishorty said:
I'm currently going through the NSGHS 2009 mathematics 2 unit trial and got stuck at question 9d,ii. I've looked at the solutions and don't understand the working out for the question. It involves find the integral of lnx/x^2 from e to 1, using the result in the previous part which was (1-lnx)/x^2.

The solution says that Inx/x^2 = -[(1-lnx)/x^2 - 1/x^2] and uses this to carry out the integration, but I don't understand how that equals that.

Can someone please help me understand it or give me another way of doing the question!

Thanks

hmm.. unsure of the actual question but I think this is the gist of it:

1.
$\begin{align*}\frac{\mathrm{d} }{\mathrm{d} x} \left ( \frac{\ln (x)}{x} \right ) &= \frac{x\times\frac{1}{x} - 1\times\ln(x)}{x^2}\\&=\frac{1-\ln(x)}{x^2}\end{align*}$

2.
$\begin{align*}\therefore \int\frac{1 - \ln (x)}{x^2}\,\mathrm{d}x &= \frac{\ln (x)}{x}\\\int\frac{1}{x^2} - \frac{\ln (x)}{x^2}\,\mathrm{d}x &= \frac{\ln (x)}{x}\\-\int\frac{\ln (x)}{x^2}\,\mathrm{d}x &= \frac{\ln (x)}{x} - \int\frac{1}{x^2}\,\mathrm{d}x\\&=\frac{\ln (x)}{x} + \frac{1}{x}\\\therefore \int\frac{\ln (x)}{x^2}\,\mathrm{d}x&=-\frac{\ln (x)}{x} - \frac{1}{x} \ (\,+\,C)\end{align*}$

My Name is Khan · Jul 2, 2012

love latex

PS: http://www.hahgay.com/

rolpsy said:
hmm.. unsure of the actual question but I think this is the gist of it:

1.
$\begin{align*}\frac{\mathrm{d} }{\mathrm{d} x} \left ( \frac{\ln (x)}{x} \right ) &= \frac{x\times\frac{1}{x} - 1\times\ln(x)}{x^2}\\&=\frac{1-\ln(x)}{x^2}\end{align*}$

2.
$\begin{align*}\therefore \int\frac{1 - \ln (x)}{x^2}\,\mathrm{d}x &= \frac{\ln (x)}{x}\\\int\frac{1}{x^2} - \frac{\ln (x)}{x^2}\,\mathrm{d}x &= \frac{\ln (x)}{x}\\-\int\frac{\ln (x)}{x^2}\,\mathrm{d}x &= \frac{\ln (x)}{x} - \int\frac{1}{x^2}\,\mathrm{d}x\\&=\frac{\ln (x)}{x} + \frac{1}{x}\\\therefore \int\frac{\ln (x)}{x^2}\,\mathrm{d}x&=-\frac{\ln (x)}{x} - \frac{1}{x} \ (\,+\,C)\end{align*}$

pokka · Jul 2, 2012

iishorty · Jul 2, 2012

Thanks rolpsy.

I'm surprised that I couldn't think of that.

That really helped, I feel so stupid now

Intergration help! (1 Viewer)

iishorty

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pokka

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