Correlation
Correlation defines a relationship between two independent variables. It tells the degree to which variables move in relation to each other. When two sets of data are related to each other, there is a correlation between them.
Linear Correlation
A correlation is used to determine the relationships between numerical and categorical variables. In other words, it is an indicator of how things are connected to one another. The correlation analysis is the study of how variables are related.
Regression Analysis
Regression analysis is a statistical method in which it estimates the relationship between a dependent variable and one or more independent variable. In simple terms dependent variable is called as outcome variable and independent variable is called as predictors. Regression analysis is one of the methods to find the trends in data. The independent variable used in Regression analysis is named Predictor variable. It offers data of an associated dependent variable regarding a particular outcome.
Evaluate
![### Integral Evaluation Example
Consider the following integral:
\[ \int \dfrac{x^2}{x^2 - 16} \, dx \]
This is a rational function where the numerator is \(x^2\) and the denominator is \(x^2 - 16\). Solving integrals of this kind typically involve algebraic manipulation such as completing the square, partial fraction decomposition, or other integral techniques.
#### Steps to Evaluate:
1. **Identify the integral:**
The given integral is:
\[ \int \dfrac{x^2}{x^2 - 16} \, dx \]
2. **Simplify:**
Notice that the denominator \(x^2 - 16\) can be factored as a difference of squares:
\[ x^2 - 16 = (x+4)(x-4) \]
3. **Express the integrand in a different form:**
It's often helpful to rewrite the integrand for partial fraction decomposition or other methods. Here, observe that:
\[ \dfrac{x^2}{x^2 - 16} = \dfrac{x^2 - 16 + 16}{x^2 - 16} = \dfrac{x^2 - 16}{x^2 - 16} + \dfrac{16}{x^2 - 16} \]
Simplifying the first term on the right side, we get:
\[ = 1 + \dfrac{16}{x^2 - 16} \]
4. **Separate the integrals:**
\[ \int \dfrac{x^2}{x^2 - 16} \, dx = \int 1 \, dx + \int \dfrac{16}{x^2 - 16} \, dx \]
The first integral is straightforward, and for the second, use a substitution \(u = x^2 - 16\).
5. **Solve each integral:**
\[ \int 1 \, dx = x \]
For \( \int \dfrac{16}{x^2 - 16} \, dx \):
Let \( u = x^2 - 16 \), hence \( du = 2x \, dx \). This substitution can help simplify and solve the integral.
Further steps would involve detailed substitution and solving each](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F715a84de-dd1a-49bc-bb04-088748385741%2Fd747fc28-5db8-49da-b392-0f7e56b4e4c0%2F8nb31mg_processed.png&w=3840&q=75)
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