Hi,

I have to agree also that differentiating the 'three' seems to be the easiest method, along with changing the form a little.
I also have to admit that i'd have to look this up myself too as i dont remember how to integrate that directly either, becoming too dependent on software these days i guess
If anyone wants to attempt the actual integration that would be interesting as well. That would provide the OP with a comparison of difficulties also.

 

If anyone wants to attempt the actual integration that would be interesting as well. That would provide the OP with a comparison of difficulties also.

\(
\int{\frac{\(x-x^3\)^{\frac{1}{3}}}{x^4}}dx
\)

This is in the family of functional forms that lends itself to finding a u-substitution that allows you to use

\(
\int{u^n}du
\)

The obvious u-substitution is

\(
u = \(x - x^3\)
du = \(1 - 3x^2) dx
\)

which isn't a very good match. Having two terms in the expression for du is problematic, but we can eliminate this using one of two alternatives:

\(
u = \frac{\(x - x^3\)}{x} = \(1 - x^2\)
du = \(2x\) dx
\)

To use this form, we would need to bring x^(1/3) out of the x^4 into the cube root, which makes the denominator messy and will not lead us to an x·dx form for du.

The other alternative is

\(
u = \frac{\(x - x^3\)}{x^3} = \(\frac{1}{x^2} - 1\)
du = -\frac{2}{x^3}dx
\)

To use this form, we would need to bring x out of the x^4 into the cube root, which leaves us with x^3 in the denominator of what is left which is just what we want. So this looks promising.

\(
\int{\frac{\(x-x^3\)^{\frac{1}{3}}}{x^4}}dx
\;
\int{\frac{\(\frac{x-x^3}{x^3}\)^{\frac{1}{3}}}{x^3}}dx
\;
\int{\(\frac{1}{x^2}-1\)^{\frac{1}{3}}\(\frac{1}{x^3}\)}dx
\;
-\frac{1}{2}\int{\(\frac{1}{x^2}-1\)^{\frac{1}{3}}\(-\frac{2}{x^3}\)}dx
\;
-\frac{1}{2}\int{u^{\frac{1}{3}}}du
\;
-\(\frac{1}{2}\)\(\frac{3}{4}\)u^{\frac{4}{3}}+C
\;
-\frac{3}{8}{\(\frac{1}{x^2} - 1\)}^{\frac{4}{3}}+C
\)

 

Keeping in mind the general relationship..

\(\frac {d\(\frac {1}{tan^{m}\(\theta\)}\)}{d\theta}=-m\(\frac {cos^{m-1}\(\theta \)}{sin^{m+1}\(\theta \) \)\)

One can make the substitution in the integral

\( x=\sin \( \theta \)\)

which after some manipulation leads to

\( \int { \frac {cos^{ \frac {5}{3}} \( \theta \)}{ sin^{\frac {11}{3}}\( \theta \)}d\theta}\)

This would then lead to a solution in the dummy variable

\( -\frac {3}{8} \( \frac {1}{tan^{\frac {8}{3}} \( \theta \)} \) + C\)

After reverting to the original variable x, this yields the solution

\( \frac {3}{8} \frac {\(x^2-1 \)\(x-x^3\)^{\frac{1}{3}}}{x^3}+C\)

Which is that obtained from the Wolfram online integrator...

 

Keeping in mind the general relationship..

\(\frac {d\(\frac {1}{tan^{m}\(\theta\)}\)}{d\theta}=-m\(\frac {cos^{m-1}\(\theta \)}{sin^{m+1}\(\theta \) \)\)

One can make the substitution in the integral

\( x=\sin \( \theta \)\)

which after some manipulation leads to

\( \int { \frac {cos^{ \frac {5}{3}} \( \theta \)}{ sin^{\frac {11}{3}}\( \theta \)}d\theta}\)

This would then lead to a solution in the dummy variable

\( -\frac {3}{8} \( \frac {1}{tan^{\frac {8}{3}} \( \theta \)} \) + C\)

After reverting to the original variable x, this yields the solution

\( \frac {3}{8} \frac {\(x^2-1 \)\(x-x^3\)^{\frac{1}{3}}}{x^3}+C\)

Which is that obtained from the Wolfram online integrator...

Boy, that's certainly an approach that would never have come to my mind on an exam -- and there's no way I would have those relationships "kept in mind".

Though you still aren't done because you have to still massage this into the form of the given answers (which isn't too bad -- only a couple lines).

But it's definitely interesting to see different ways of tackling it.

 

@t_n_k
Show off! I mean this in the most complementary way possible.

 

It is interesting that one might naturally assume the original question requires one to perform the integration.
As suggested earlier, why not simply look at the answer options (A) to (D) provided, eliminate any cases which are clearly wrong and, if necessary, differentiate the remainder to arrive at a conclusion. I would think this a perfectly valid approach given the wording of the question.

 

Especially, since it is well know by students of the calculus, that differentiation is far more likely to succeed by the application of purely mechanical methods than integration, as WBahn and t_n_k have so ably demonstrated.

 

Hi,

Nice solutions guys. This helps show the OP what would be involved for finding the result directly.

I have a feeling this might be part of the course where they are teaching the integral as an 'antiderivative'.