Integration using completing the square and the derivative of arctan(x) | Khan Academy
TLDRThe video script presents a step-by-step guide to finding the indefinite integral of a complex rational function. The process involves simplifying the denominator, completing the square to resemble the derivative of arctan, and employing u-substitution to transform the integral into a form involving arctan. The final result is an expression involving the arctangent function, showcasing the power of integration techniques in solving mathematical problems.
Takeaways
- π The problem involves finding the indefinite integral of a complex rational function.
- π€ The initial approach may involve trying multiple integration techniques but facing difficulties.
- π The strategy is to complete the square in the denominator to transform the expression.
- π By completing the square, the expression is reformulated to resemble the derivative of arctan.
- π’ The first step is to simplify the denominator by factoring out a common coefficient.
- 𧩠Adding and subtracting the same value (in this case, 9) helps create a perfect square trinomial.
- π The integral can be rewritten with the perfect square and an additional term, resembling a standard form.
- π A u-substitution is introduced to simplify the expression further, making the arctan connection more evident.
- π± The substitution involves setting u as x - 3/2 and du as 1/2 dx, which simplifies the integral.
- π€ After the u-substitution, the integral simplifies to 1/10 times the arctan of u.
- π The final result is the indefinite integral expressed as 1/10 times the arctan of (x - 3/2) plus a constant C.
Q & A
What is the given integral that the instructor is attempting to solve?
-The given integral is one over five x squared minus 30x plus 65 dx.
What is the first step the instructor suggests in solving this integral?
-The first step is to simplify the denominator by factoring out a 5 to make the coefficient of the x squared term equal to 1.
How does the instructor plan to transform the given integral to make it resemble the derivative of arctan?
-The instructor plans to complete the square in the denominator to get it in a form that resembles the derivative of arctan.
What is the significance of completing the square in this context?
-Completing the square helps to transform the integral into a form that looks like the derivative of the arctan function, which simplifies the process of integration.
What is the role of the coefficient of x in the process of completing the square?
-The coefficient of x, which is -6 in this case, is used to determine the value to be added and subtracted (half of the coefficient squared) to complete the square.
How does the instructor perform u-substitution in this integral?
-The instructor sets u equal to x minus three over two, with du equal to one half dx, and then substitutes these into the integral to simplify it further.
What does the derivative of arctan of u look like?
-The derivative of arctan of u is one over u squared plus one.
What is the final result of the integral after performing the u-substitution and simplification?
-The final result is 1/10 times the arctan of (x minus three over two) plus a constant C, representing the indefinite integral.
Why is it important to include the constant C in the final result?
-The constant C is included because we are solving an indefinite integral, which represents a family of functions that differ by a constant.
How does the instructor ensure that the value of the denominator does not change when completing the square?
-The instructor adds and subtracts the same value (in this case, 9) to complete the square without changing the value of the denominator.
What is the final expression of the integral after the instructor completes the square and performs the u-substitution?
-The final expression is 1/10 times the arc tangent of (x minus three over two) plus a constant C.
Outlines
π Solving Indefinite Integral with Complete Square Method
This paragraph delves into the process of finding the indefinite integral of a given function. The instructor introduces a complex integral expression and encourages the audience to attempt solving it before revealing the method. The approach taken involves simplifying the denominator by factoring out a number to make the coefficient of the x squared term equal to one. The instructor then employs the technique of completing the square to transform the expression into a form resembling the derivative of the arctangent function. The explanation proceeds with a step-by-step breakdown of the process, including the addition and subtraction of a value to create a perfect square and the subsequent rewriting of the integral to highlight its similarity to the arctangent function. The paragraph concludes with the introduction of a substitution method to further simplify the integral and eventually leads to the solution involving the arctangent function, emphasizing the importance of recognizing the derivative of arctangent in the process.
π Finalizing the Indefinite Integral with Reverse Substitution
In this paragraph, the solution to the indefinite integral is completed through the use of reverse substitution. The instructor starts by reiterating the integral expression derived from the previous paragraph and proceeds to replace the variable 'u' with the expression 'x - 3/2', which was determined during the substitution process. The final step involves simplifying the integral to its final form, which is the arctangent of '(x - 3)/2' multiplied by 1/10, plus a constant 'C'. This conclusion marks the end of the process, providing the audience with a clear and comprehensive solution to the original integral problem.
Mindmap
Keywords
π‘indefinite integral
π‘multiple integration techniques
π‘completing the square
π‘arctangent
π‘u-substitution
π‘derivative
π‘constant of integration
π‘reverse substitution
π‘quadratic expression
π‘coefficient
π‘perfect square
Highlights
The problem presented is finding the indefinite integral of a complex rational function.
The approach involves multiple integration techniques and completing the square in the denominator.
The goal is to transform the given expression into a form that resembles the derivative of arctangent.
The first step is to simplify the denominator by factoring out a coefficient.
Completing the square is necessary to transform the expression into a perfect square trinomial.
The process includes adding and subtracting the same value to maintain the integrity of the expression.
The integral is rewritten with a perfect square and an additional term, resembling a form related to arc tangent.
An u-substitution is introduced to simplify the integral and reveal its relation to the derivative of arc tangent.
The substitution involves setting u equal to a specific expression involving x and simplifying the integral accordingly.
The derivative of arc tangent is recognized as a key component in the integral.
The integral is expressed in terms of the arc tangent function with a coefficient and constant.
The reverse substitution is performed to express the final result in terms of the original variable x.
The final answer is the integral as a function of arc tangent with the original variable x.
The process demonstrates a methodical approach to solving complex integrals through algebraic manipulation and substitution.
The video provides a detailed walkthrough of the steps, making it an educational resource for similar problems.
The solution showcases the application of advanced calculus techniques in integral calculus.
Transcripts
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