Evaluate The Following Integrals: 1. ∫ (x^4)/(x-1) Dx 2. ∫ (3t-2)/(t+1) Dt 3. ∫ (5x+1)/((2x+1)(x-1)) Dx 4. ∫ Y/((y+4)(2y-1)) Dy 5. ∫₀¹ 2/(2x²+3x+1) Dx 6. ∫₀¹ (x-4)/(x²-5x+6) Dx

by ADMIN 177 views

In the realm of calculus, integral evaluation stands as a cornerstone, enabling us to determine areas, volumes, and a myriad of other quantities. This article delves into the methods for evaluating various integrals, presenting detailed solutions and explanations to enhance your understanding. We'll explore polynomial division, partial fraction decomposition, and definite integrals, equipping you with the tools to tackle complex problems.

1. ∫ (x^4)/(x-1) dx

To evaluate the integral ∫ (x^4)/(x-1) dx, we first perform polynomial long division. This allows us to rewrite the integrand as a sum of simpler terms, making the integration process more manageable. Polynomial long division is a crucial technique in calculus, particularly when dealing with rational functions where the degree of the numerator is greater than or equal to the degree of the denominator. By dividing x^4 by (x-1), we obtain a quotient and a remainder that simplify the original expression.

The division yields:

x^4 = (x-1)(x^3 + x^2 + x + 1) + 1

Thus, (x^4)/(x-1) can be expressed as:

(x^4)/(x-1) = x^3 + x^2 + x + 1 + 1/(x-1)

Now, the integral becomes:

∫ (x^4)/(x-1) dx = ∫ (x^3 + x^2 + x + 1 + 1/(x-1)) dx

This integral can be solved term by term using the power rule and the logarithmic rule for integration. The power rule states that ∫x^n dx = (x^(n+1))/(n+1) + C, while the integral of 1/(x-1) is ln|x-1| + C. Applying these rules, we get:

∫ x^3 dx = (x^4)/4 + C ∫ x^2 dx = (x^3)/3 + C ∫ x dx = (x^2)/2 + C ∫ 1 dx = x + C ∫ 1/(x-1) dx = ln|x-1| + C

Combining these results, we find:

∫ (x^4)/(x-1) dx = (x^4)/4 + (x^3)/3 + (x^2)/2 + x + ln|x-1| + C

This comprehensive solution showcases the power of polynomial long division in simplifying complex rational functions, enabling us to evaluate their integrals effectively. Remember, the constant of integration, C, is a crucial part of the final answer, representing the family of antiderivatives.

2. ∫ (3t-2)/(t+1) dt

To tackle the integral ∫ (3t-2)/(t+1) dt, we again employ polynomial division, a fundamental technique for simplifying rational functions. In this case, dividing (3t-2) by (t+1) helps us rewrite the integrand in a more manageable form. This is particularly useful when the degree of the numerator is equal to or greater than the degree of the denominator. By performing this division, we can break down the complex fraction into simpler terms that are easier to integrate.

The division gives us:

3t - 2 = 3(t + 1) - 5

So, (3t-2)/(t+1) can be rewritten as:

(3t-2)/(t+1) = 3 - 5/(t+1)

Now, the integral becomes:

∫ (3t-2)/(t+1) dt = ∫ (3 - 5/(t+1)) dt

This integral is now a sum of two simpler integrals: the integral of a constant and the integral of a fraction with a linear denominator. The integral of a constant is straightforward, and the integral of 5/(t+1) can be solved using a simple substitution or recognizing it as a logarithmic form.

We can break it down further:

∫ 3 dt = 3t + C ∫ 5/(t+1) dt = 5 ∫ 1/(t+1) dt

The integral of 1/(t+1) is ln|t+1| + C. Therefore,

5 ∫ 1/(t+1) dt = 5ln|t+1| + C

Combining these results, we have:

∫ (3t-2)/(t+1) dt = 3t - 5ln|t+1| + C

This detailed solution illustrates how polynomial division transforms a seemingly complex integral into a straightforward problem. The ability to manipulate rational functions in this way is a key skill in integral calculus, allowing for efficient and accurate solutions. Don't forget the constant of integration, C, which accounts for the family of possible antiderivatives.

3. ∫ (5x+1)/((2x+1)(x-1)) dx

The integral ∫ (5x+1)/((2x+1)(x-1)) dx requires the technique of partial fraction decomposition. This method is crucial for integrating rational functions where the denominator can be factored into distinct linear factors. Partial fraction decomposition allows us to break down a complex fraction into a sum of simpler fractions, each of which is easier to integrate. This is a fundamental technique in calculus, especially when dealing with rational functions in integration problems.

First, we decompose the fraction:

(5x+1)/((2x+1)(x-1)) = A/(2x+1) + B/(x-1)

To find the constants A and B, we multiply both sides by the denominator (2x+1)(x-1):

5x + 1 = A(x-1) + B(2x+1)

Now, we can solve for A and B by substituting suitable values for x. Let's start with x = 1:

5(1) + 1 = A(1-1) + B(2(1)+1) 6 = 0 + 3B B = 2

Next, let's use x = -1/2:

5(-1/2) + 1 = A(-1/2 - 1) + B(2(-1/2) + 1) -5/2 + 1 = A(-3/2) + 0 -3/2 = -3/2 A A = 1

So, we have A = 1 and B = 2. The integral becomes:

∫ (5x+1)/((2x+1)(x-1)) dx = ∫ (1/(2x+1) + 2/(x-1)) dx

Now, we can integrate each term separately:

∫ 1/(2x+1) dx = (1/2)ln|2x+1| + C ∫ 2/(x-1) dx = 2ln|x-1| + C

Combining these results, we get:

∫ (5x+1)/((2x+1)(x-1)) dx = (1/2)ln|2x+1| + 2ln|x-1| + C

This detailed walkthrough illustrates the power and elegance of partial fraction decomposition. By breaking down the complex rational function, we were able to integrate it easily. This technique is indispensable for solving a wide range of integrals involving rational functions. Remember to always include the constant of integration, C, in your final answer.

4. ∫ y/((y+4)(2y-1)) dy

To evaluate the integral ∫ y/((y+4)(2y-1)) dy, we again utilize the powerful technique of partial fraction decomposition. This method is essential for breaking down rational functions into simpler components that can be easily integrated. When the denominator of a rational function can be factored, partial fraction decomposition provides a systematic way to express the original fraction as a sum of simpler fractions, each corresponding to a factor in the denominator.

We begin by decomposing the fraction:

y/((y+4)(2y-1)) = A/(y+4) + B/(2y-1)

To find the constants A and B, we multiply both sides by the denominator (y+4)(2y-1):

y = A(2y-1) + B(y+4)

Now, we solve for A and B by substituting suitable values for y. Let's start with y = -4:

-4 = A(2(-4)-1) + B(-4+4) -4 = -9A A = 4/9

Next, let's use y = 1/2:

1/2 = A(2(1/2)-1) + B(1/2 + 4) 1/2 = 0 + (9/2)B B = 1/9

So, we have A = 4/9 and B = 1/9. The integral becomes:

∫ y/((y+4)(2y-1)) dy = ∫ (4/9)/(y+4) + (1/9)/(2y-1) dy

Now, we integrate each term separately:

∫ (4/9)/(y+4) dy = (4/9)ln|y+4| + C ∫ (1/9)/(2y-1) dy = (1/18)ln|2y-1| + C

Combining these results, we get:

∫ y/((y+4)(2y-1)) dy = (4/9)ln|y+4| + (1/18)ln|2y-1| + C

This detailed solution demonstrates the effectiveness of partial fraction decomposition in simplifying complex integrals. By breaking down the rational function, we transformed the integral into a sum of simpler integrals that are easily evaluated. This technique is a cornerstone of integral calculus and is applicable to a wide range of problems. Remember to include the constant of integration, C, in your final answer.

5. ∫₀¹ 2/(2x²+3x+1) dx

To evaluate the definite integral ∫₀¹ 2/(2x²+3x+1) dx, we combine the techniques of partial fraction decomposition and definite integration. First, we need to decompose the rational function 2/(2x²+3x+1) into simpler fractions. This involves factoring the denominator and expressing the original fraction as a sum of fractions with linear denominators.

The denominator can be factored as:

2x² + 3x + 1 = (2x+1)(x+1)

Now, we decompose the fraction:

2/((2x+1)(x+1)) = A/(2x+1) + B/(x+1)

To find the constants A and B, we multiply both sides by the denominator (2x+1)(x+1):

2 = A(x+1) + B(2x+1)

Now, we solve for A and B by substituting suitable values for x. Let's start with x = -1:

2 = A(-1+1) + B(2(-1)+1) 2 = -B B = -2

Next, let's use x = -1/2:

2 = A(-1/2 + 1) + B(2(-1/2) + 1) 2 = (1/2)A A = 4

So, we have A = 4 and B = -2. The integral becomes:

∫₀¹ 2/(2x²+3x+1) dx = ∫₀¹ (4/(2x+1) - 2/(x+1)) dx

Now, we integrate each term separately:

∫ 4/(2x+1) dx = 2ln|2x+1| + C ∫ -2/(x+1) dx = -2ln|x+1| + C

Combining these results, we get:

∫ (4/(2x+1) - 2/(x+1)) dx = 2ln|2x+1| - 2ln|x+1| + C

Now, we evaluate the definite integral from 0 to 1:

[2ln|2x+1| - 2ln|x+1|]₀¹ = (2ln|2(1)+1| - 2ln|1+1|) - (2ln|2(0)+1| - 2ln|0+1|) = (2ln(3) - 2ln(2)) - (2ln(1) - 2ln(1)) = 2ln(3) - 2ln(2) = 2(ln(3) - ln(2)) = 2ln(3/2)

This detailed solution combines partial fraction decomposition with the evaluation of a definite integral. By breaking down the rational function and then applying the limits of integration, we arrived at the final answer. This approach is essential for solving many calculus problems involving rational functions and definite integrals.

6. ∫₀¹ (x-4)/(x²-5x+6) dx

To evaluate the definite integral ∫₀¹ (x-4)/(x²-5x+6) dx, we again employ the technique of partial fraction decomposition, a fundamental tool for integrating rational functions. This method involves breaking down the integrand into simpler fractions that are easier to integrate. In this case, we need to factor the denominator and then express the original fraction as a sum of simpler fractions.

First, we factor the denominator:

x² - 5x + 6 = (x-2)(x-3)

Now, we decompose the fraction:

(x-4)/((x-2)(x-3)) = A/(x-2) + B/(x-3)

To find the constants A and B, we multiply both sides by the denominator (x-2)(x-3):

x - 4 = A(x-3) + B(x-2)

Now, we solve for A and B by substituting suitable values for x. Let's start with x = 2:

2 - 4 = A(2-3) + B(2-2) -2 = -A A = 2

Next, let's use x = 3:

3 - 4 = A(3-3) + B(3-2) -1 = B B = -1

So, we have A = 2 and B = -1. The integral becomes:

∫₀¹ (x-4)/(x²-5x+6) dx = ∫₀¹ (2/(x-2) - 1/(x-3)) dx

Now, we integrate each term separately:

∫ 2/(x-2) dx = 2ln|x-2| + C ∫ -1/(x-3) dx = -ln|x-3| + C

Combining these results, we get:

∫ (2/(x-2) - 1/(x-3)) dx = 2ln|x-2| - ln|x-3| + C

Now, we evaluate the definite integral from 0 to 1:

[2ln|x-2| - ln|x-3|]₀¹ = (2ln|1-2| - ln|1-3|) - (2ln|0-2| - ln|0-3|) = (2ln(1) - ln(2)) - (2ln(2) - ln(3)) = (0 - ln(2)) - (2ln(2) - ln(3)) = -ln(2) - 2ln(2) + ln(3) = -3ln(2) + ln(3) = ln(3) - ln(2³) = ln(3) - ln(8) = ln(3/8)

This detailed solution showcases the application of partial fraction decomposition in evaluating a definite integral. By breaking down the rational function and then applying the limits of integration, we were able to find the precise value of the integral. This approach is vital for solving a wide array of calculus problems involving rational functions and definite integrals.

In this comprehensive guide, we've explored various techniques for evaluating integrals, including polynomial division, partial fraction decomposition, and definite integration. Each method provides a unique approach to simplifying complex integrals and arriving at accurate solutions. By mastering these techniques, you'll be well-equipped to tackle a wide range of calculus problems and deepen your understanding of integral calculus.