Evaluating the Integral of ln(1+t)t^3/(1+t)

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Homework Statement


\int \left( \dfrac{ln(1+\sqrt[6]{x})}{\sqrt[3]{x} + \sqrt{x}} \right) dx

The Attempt at a Solution


Let x^(1/6) = t

\int \dfrac{ln(1+t)t^3}{1+t} dt
 
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You want to evaluate:
$$\int \left( \dfrac{\ln(1+\sqrt[6]{x})}{\sqrt[3]{x} + \sqrt{x}} \right) dx$$

You attempted it by doing the substitution: ##x^{(1/6)} = t## - which gets you:

$$\int \dfrac{\ln(1+t)t^3}{1+t} dt$$
... and, from there, you get stuck?

Are you missing a factor of 6 in there?

Have you tried putting u=1+t ?
 
Simon Bridge said:
You want to evaluate:
$$\int \left( \dfrac{\ln(1+\sqrt[6]{x})}{\sqrt[3]{x} + \sqrt{x}} \right) dx$$

You attempted it by doing the substitution: ##x^{(1/6)} = t## - which gets you:

$$\int \dfrac{\ln(1+t)t^3}{1+t} dt$$
... and, from there, you get stuck?

Are you missing a factor of 6 in there?

Have you tried putting u=1+t ?

I forgot that 6. By the way it won't make my life easier. I tried substituting z=1+t as well and came up with this.

\displaystyle \int \dfrac{lnz(z-1)^3}{z} dz

Also, how do you get your integral sign bigger? Mine renders as small when I use \int. Here I've used \displaystyle to get it bigger.
 
Last edited:
If you want to know how I get a particular format for something, use the "quote" button at the bottom of my posts - it shows you the markup.

Your int signs render small because you are using in-line math style - the "itex" or double-hash tags. To get things to work better, use the displaymath style - the "tex" or double-dollar tags.

Similarly, you can get standard functions to render properly by putting a backslash in front of them ... like \ln for the natural logarithm.

After the substitution you should have something of form ##\int f(z)\ln|z|\; dz## ...

Arm yourself with a bunch of tables to help your strategy:
http://en.wikipedia.org/wiki/List_of_integrals_of_logarithmic_functions
http://en.wikipedia.org/wiki/List_of_logarithmic_identities
 
Or you can use partial integration to massage the logarithm out (although things will get slightly messy)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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