How to Integrate the Indefinite Integral of (2-y)/(1-y) Using Different Techniques

This article will guide you through the process of integrating the indefinite integral of the expression ∫ (2-y)/(1-y) dy using various mathematical techniques. We will present two different methods to arrive at the solution, including substitution and algebraic simplification.

Method 1: Algebraic Simplification and Integration

Let's start with the expression ∫ (2-y)/(1-y) dy. We can simplify the integrand using algebraic manipulation. First, we rewrite the integrand as follows:

Step 1: Rewrite the expression

∫ (1-y 1)/(1-y) dy ∫ (1-y)/(1-y) dy ∫ (1)/(1-y) dy

Notice that the integrand can be split into two simpler integrals:

Step 2: Split the integral

∫ (1-y)/(1-y) dy ∫ (1)/(1-y) dy ∫ 1 dy ∫ (1)/(1-y) dy

Note that 1/(1-y) -1/(y-1). Let's integrate each term separately.

Step 3: Integrate each term

∫ 1 dy ∫ (1)/(1-y) dy y - ln|1-y| C

Here, the first term integrates to y and the second term integrates to -ln|1-y| (remember, the absolute value is necessary because the natural logarithm is only defined for positive values).

Method 2: Substitution Method

Another way to approach the integral is by using the substitution method. Let's follow this approach step-by-step:

Step 1: Define the substitution

Let 1-y u so that -dy du → dy -du

Now, substitute the value of u and dy into the integral:

Step 2: Substitute and simplify

∫ (2-y)/(1-y) dy -∫ (2-u)/u du

Since 2-u 1-u 1, we can further simplify the integrand:

-∫ (1 1-u)/u du -∫ (1/u) du - ∫ (1) du

The first integral is the integral of 1/u, which is -ln|u|, and the second integral is simply -1. Therefore, we integrate each term:

Step 3: Integrate and simplify further

-∫ (1/u) du - ∫ (1) du -ln|u| - u C

Substitute back u 1-y:

-ln|1-y| - (1-y) C y - ln|1-y| C

This confirms our result from the first method: y - ln|1-y| C.

Conclusion

Both methods lead to the same result, confirming the correctness of our calculations. The integral of the expression (2-y)/(1-y) is y - ln|1-y| C. This approach not only demonstrates the power of algebraic manipulation but also the versatility of integration techniques. Mastering these methods can be incredibly helpful in solving complex integrals in calculus.

Key Takeaways:

Algebraic simplification can greatly facilitate integration. The substitution method is a powerful tool, especially when dealing with more complex expressions. Absolute values are crucial in integration involving natural logarithms.

To further your understanding of these concepts, you can practice more problems and explore applications of these techniques in real-world scenarios. For those interested in delving deeper, you might consider exploring advanced integration techniques such as integration by parts or partial fractions.