Solving the Integral of x^5/sqrt(1 - x^3) with Steps and Methods

In this article, we will explore the process of solving the integral of x^5/sqrt(1 - x^3). This type of integral requires a deep understanding of various integration techniques, including substitution. Let's dive into the detailed steps and methods to solve this problem.

Step 1: Initial Setup

The given integral is:

I ∫(x^5/sqrt(1 - x^3)) dx

To tackle this, we will start by undergoing an initial transformation to simplify the given function:

I ∫(x^5/sqrt(1 - x^3)) dx ∫(x^2 1 - x^2)/sqrt(1 - x^3) dx

This simplifies to:

I ∫(x^2 1)/sqrt(1 - x^3) dx - ∫(x^2)/sqrt(1 - x^3) dx

Step 2: Simplifying Further

Next, we will separate the integral into two parts:

I ∫(x^2 1)/sqrt(1 - x^3) dx - ∫(x^2)/sqrt(1 - x^3) dx

Let us work on each integral separately starting with the first one:

I1 ∫(x^2 1)/sqrt(1 - x^3) dx

Notice that the numerator is 1 less than the square root term. We can thus split it as:

I1 ∫(1 - (1 - x^3 - 1))/sqrt(1 - x^3) dx

The term (1 - x^3 - 1) simplifies to -x^3, thus:

I1 ∫(1 - x^3 - 1)/sqrt(1 - x^3) dx

Step 3: Integration by Substitution

For the first integral I1, we use the substitution:

u 1 - x^3 du -3x^2 dx

So, the integral becomes:

I1 -1/3 ∫ 1/u^(1/2) du

Integrating gives us:

I1 -2/9 (1 - x^3)^(3/2) C1

For the second integral I2:

I2 ∫ x^2 / (1 - x^3)^(1/2) dx

Again, using the substitution:

u 1 - x^3 du -3x^2 dx

So, the integral becomes:

I2 -1/3 ∫ 1/u^(1/2) du

Integrating gives us:

I2 -2/3√(1 - x^3) C2

Combining I1 and I2, we have:

I -2/9 (1 - x^3)^(3/2) 2/3√(1 - x^3) C

Final Answer:

I -2/9 (1 - x^3)^(3/2) 2/3√(1 - x^3) C

Conclusion

By breaking down the integral into smaller parts and using the appropriate substitution, we were able to solve the given integral step by step. This approach demonstrates the power and effectiveness of integration by substitution in solving complex mathematical problems.

Additional Resources:

For further learning and practice, visit the following pages:

Integration Techniques: Substitution Methods: Advanced Calculus: