Evaluate the integral. ∫(x)/(x^2+6 x+10) d x

Name: Problem 41, Chapter 9: Calculus A New Horizon
Uploaded: 2021-12-30T16:47:03-08:00
Duration: 6 min 22 s
Channel: Uma Kumari
Description: evaluate the integral. ∫(x)/(x^2+6 x+10) d x

Key Concepts

Integration Involving Logarithms

This concept refers to the occurrence of logarithmic functions as antiderivatives, particularly when integrating rational functions that, after appropriate substitutions, yield integrals of the form f'(x)/f(x). Recognizing this structure allows one to directly integrate to obtain a logarithmic function as part of the solution.

u-Substitution

u-Substitution is an integration technique where a substitution is chosen to simplify the integrand by setting u equal to a part of the expression. This method is especially useful when the derivative of the chosen u appears elsewhere in the integrand, allowing the integral to be rewritten in a simpler form.

Completing the Square

Completing the square is a method for rewriting a quadratic expression in the form of a perfect square plus a constant. This transformation is particularly useful in integration because it can simplify the quadratic in the denominator, making it easier to apply standard integration techniques or substitutions.

Integration of Rational Functions

This concept involves methods for integrating expressions where both the numerator and the denominator are polynomials. Techniques such as algebraic manipulation, substitution, and partial fractions are often employed to simplify these integrals into forms that are easier to evaluate.

Transcript

00:01 In this question we are required to find the value of integration under root 3 minus 2x minus x square d x.

00:13 So let's see how to solve this question.

00:16 First of all, let's make a whole square in this square root.

00:20 So we will have integration under root of we are adding and subtracting 1 in this expression.

00:28 So we will have 4 minus x squared.

00:32 Square minus 2x minus 1 d x this will be equals to integration under root of 4 minus x plus 1 to the power 2 d x now to solve this integral consider x plus 1 is equal to 2 sine theta therefore by the differentiation we can write d x is equal to 2 cosine theta d theta and now substitute all these values in the above integral so we will have integration under root 4 minus x plus 1 to the power 2 d x is equals to integration under root of 4 minus 4 sine square theta multiplied by d x that means 2 cosine theta d theta this will be equals to 4 integration under root of 1 minus sine square theta multiplied by cosine theta d theta we know that sine square theta plus cosine square theta is equals to 1 therefore under root of 1 minus sine square theta will be equals to cosine theta now substitute this value in the above integral so integral will be equals to 4 integration cosine square theta d theta and we know that cosine 2 theta is equals to 2 cosine square theta minus 1 therefore cosine square theta will be equals to 1 upon 2 1 plus 1 plus cosine 2 theta.

02:53 Now substitute this value in the above integral, so we can write integration under root 4 minus x plus 1 to the power 2, d x is equals to 4 upon 2 integration 1 plus cosine 2 theta d theta.

03:19 Now let's integrate each term with respect to theta so we will have 2 into the integration of 1 d theta will be equals to theta plus the integration of cosine 2 theta d theta will be equals to sine 2 theta upon 2 when we further calculate this we get 2 theta plus 2 sine theta into cosine theta plus a constant of integration in the beginning of problem we have considered x plus 1 1 is equal to 2 sine theta...

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