Integration by Parts: Solving Complex Functions like ( int x^3 ln x , dx )

In calculus, integration by parts is a technique used to integrate the product of two functions. It is based on the product rule of differentiation. This technique is particularly useful for solving integrals of products where one function can be integrated and the other can be differentiated repeatedly.

Introduction to Integration by Parts

Integration by parts is derived from the product rule of differentiation. The formula for integration by parts is given by:

Rule:
$$int u , dv uv - int v , du$$

Here, (u) and (dv) are chosen partially from the given integral. The key to success in integration by parts is selecting (u) and (dv) wisely. Typically, we select (u) as the function that become simpler upon differentiation, and (dv) as the function that is easy to integrate.

Applying Integration by Parts to Solve ( int x^3 ln x , dx )

Let's consider the integral:

I ( int x^3 ln x , dx )

To apply integration by parts, we need to choose (u) and (dv) as follows:

Step 1: Choose (u) and (dv)
Let:
u (ln x) and
dv (x^3 , dx)

Step 2: Calculate (du) and (v)
Then,
du (frac{1}{x} dx)
and
v (frac{x^4}{4})

Step 3: Apply the integration by parts formula

Using the formula (int u , dv uv - int v , du):

I (ln x cdot frac{x^4}{4} - int frac{x^4}{4} cdot frac{1}{x} , dx)

Simplify the integral:

I (frac{x^4 ln x}{4} - int frac{x^3}{4} , dx)

Now, integrate (int frac{x^3}{4} , dx):

(int frac{x^3}{4} , dx frac{1}{4} int x^3 , dx frac{1}{4} cdot frac{x^4}{4} frac{x^4}{16})

Therefore:

(I frac{x^4 ln x}{4} - frac{x^4}{16} C)

Here, (C) is the constant of integration.

Verification and Alternate Solutions

Another way to approach this integral is by using a different choice of (u) and (dv). Let:

u (x^3) and dv (ln x , dx)

Then, du (3x^2 dx)

Using the integration by parts formula:

I (x^3 ln x - int 3x^2 ln x , dx)

Continue this process by splitting the integral further:

I (x^3 ln x - 3 int x^2 ln x , dx)

Repeat the integration by parts until the original integral is simplified.

For instance, if we use u (x^2) and dv (ln x , dx), repeat the process iteratively to solve the remaining integrals.

Conclusion

Integration by parts is a powerful technique to solve complex integrals. By choosing (u) and (dv) wisely, you can simplify the given integral and solve it step by step.

Remember, the key to solving integrals by parts is practice and patience. The more problems you solve, the better you will become at recognizing which functions should be chosen as (u) and (dv).

Stay tuned for more tips and techniques in calculus and other areas of mathematics!