Integrate 1/(x^1/3 + x^1/4): Can Anyone Help?

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In summary, the purpose of integrating 1/(x^1/3 + x^1/4) is to find the antiderivative or indefinite integral of the given function. This involves finding a function whose derivative is equal to 1/(x^1/3 + x^1/4). There are various techniques that can be used to solve this integral, such as substitution, partial fractions, trigonometric substitutions, and integration by parts. It is important to understand the fundamentals of integration and practice in order to successfully solve this integral and other similar problems. The applications of integrating 1/(x^1/3 + x^1/4) are vast and can be seen in various real-world problems and fields such as
  • #1
stoffer
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Can anybody help me with the integral of 1/ (x^1/3 + x^1/4)
(cube root and fourth root of x) I don't really know where to start.

Also my roomate and i were wondering if x^sin(x) exists or if it has to be expressed and integrated as some sort of series.(something i haven't learned yet)
 
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  • #2
For first substitute x=t12 which will eliminate rational power

For part2 Post it in Calculus & analysis Section
 
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  • #3


Sure, I can help you with the integral of 1/(x^1/3 + x^1/4). To start, we can use the substitution u = x^(1/12) to simplify the integral. This will result in the integral of 12/(u^4 + u^3). From here, we can use partial fractions to break down the integrand into simpler fractions and then integrate each term separately. I would suggest reviewing the steps for partial fractions if you are not familiar with it.

In regards to x^sin(x), it does exist and can be integrated as a series, but it can also be expressed using special functions such as the exponential integral or the sine integral. If you haven't learned about these yet, it may be best to stick with the series representation for now. I would also suggest consulting your textbook or a math tutor for further guidance. Hope this helps!
 

1. What is the purpose of integrating 1/(x^1/3 + x^1/4)?

The purpose of integrating 1/(x^1/3 + x^1/4) is to find the antiderivative or indefinite integral of the given function. This means finding a function whose derivative is equal to 1/(x^1/3 + x^1/4). Integration is an important concept in mathematics and is used to solve various problems in physics, engineering, and other fields.

2. How do you solve the integral of 1/(x^1/3 + x^1/4)?

To solve the integral of 1/(x^1/3 + x^1/4), you can use the substitution method or partial fractions method. In the substitution method, you substitute a new variable for the expression inside the parentheses and then solve the resulting integral. In the partial fractions method, you decompose the given function into simpler fractions and then integrate each fraction separately.

3. Are there any special techniques for integrating 1/(x^1/3 + x^1/4)?

Yes, there are special techniques such as trigonometric substitutions, integration by parts, and use of tables for integrals. These techniques can be used depending on the form of the given function. For example, trigonometric substitutions are useful for integrals involving square roots, while integration by parts is useful for integrals involving products of functions.

4. Can anyone help me with the integration of 1/(x^1/3 + x^1/4)?

Yes, there are many resources available online and in textbooks that can help you with the integration of 1/(x^1/3 + x^1/4). You can also seek help from a math tutor or your professor if you are struggling with the concept. It is important to practice and understand the fundamentals of integration in order to successfully solve this integral and other similar problems.

5. What are the applications of integrating 1/(x^1/3 + x^1/4)?

Integrating 1/(x^1/3 + x^1/4) has various applications in real-world problems. For example, it can be used to find the area under a curve, calculate the work done by a variable force, or determine the velocity of an object given its acceleration. Integration is also used in many fields such as physics, engineering, economics, and statistics to solve various problems and make predictions.

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