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

  • Thread starter utkarshakash
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In summary: ... like:\displaystyle\int_{-\infty}^\infty f(z) dy\int_{-\infty}^\infty f(z) dz+\int_{-\infty}^\infty f(z) dx... where the last two are equivalent to the first.
  • #1
utkarshakash
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Homework Statement


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

The Attempt at a Solution


Let x^(1/6) = t

[itex]\int \dfrac{ln(1+t)t^3}{1+t} dt [/itex]
 
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  • #2
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 ?
 
  • #3
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.

[itex]\displaystyle \int \dfrac{lnz(z-1)^3}{z} dz[/itex]

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:
  • #4
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
 
  • #5
Or you can use partial integration to massage the logarithm out (although things will get slightly messy)
 

What does it mean to "evaluate an integral"?

Evaluating an integral means to find the numerical value of the integral, which represents the area under a curve in a specific interval.

How do you evaluate an integral?

To evaluate an integral, you can use various techniques such as substitution, integration by parts, or trigonometric identities. First, identify the appropriate method to use, then apply the necessary steps to solve the integral.

What are the limits of integration?

Limits of integration are the upper and lower bounds that define the interval in which the integral is being evaluated. These limits can be numbers, variables, or infinity.

Can every integral be evaluated?

No, not every integral can be evaluated analytically. Some integrals are considered to be unsolvable, and in those cases, numerical methods such as the trapezoidal rule or Simpson's rule can be used to approximate the value of the integral.

Why is it important to evaluate integrals?

Evaluating integrals is crucial in many areas of science, such as physics, engineering, and economics. It allows us to calculate important quantities such as displacement, velocity, and area, which are essential in understanding real-world phenomena and making accurate predictions.

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