Evaluate the integral. ∫(x^3+3 x^2+x+9)/((x^2+1)(x^2+3)) d x...

Name: Problem 31, Chapter 7: Calculus Early Transcendentals
Uploaded: 2020-08-21T23:34:13-08:00
Duration: 5 min 39 s
Channel: Patrick Vaughn
Description: Evaluate the integral. ∫(x^3+3 x^2+x+9)/((x^2+1)(x^2+3)) d x

Key Concepts

Rational Functions

Rational functions are expressions formed by the ratio of two polynomials. In integration, these functions often require special techniques because their antiderivatives are not always obvious without manipulation, such as dividing polynomials or decomposing the expression.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to rewrite a rational function as a sum of simpler fractions whose denominators are polynomials of lower degree. This method simplifies the integration process by reducing the original fraction into a series of easier-to-integrate components.

Integration of Rational Functions

Integrating rational functions usually involves first expressing the integrand in a simpler form through algebraic techniques like polynomial division or partial fraction decomposition. Once decomposed, standard integration techniques and formulas can be applied to each term individually.

Standard Integral Forms

Standard integral forms, such as the integral of 1/(x^2 + a^2) leading to an arctan function or logarithmic forms from linear denominators, are essential tools in calculus. Familiarity with these forms allows for the quick evaluation of many integrals once a rational function has been appropriately decomposed.

Transcript

00:01 So we want to evaluate the integral of x cubed plus 3x squared plus x plus 9 divided by x squared plus 1 times x squared plus 3 dx.

00:19 So we have the numerator and the denominator are both polynomials, and the denominator has a higher order than the numerator.

00:29 So we can use partial fractions to break this up into easier fractions to deal with.

00:36 And this is going to illustrate what we do when we have fractions that aren't linear, that we can't break down into linear fractions without using imaginary roots.

00:48 So let's go ahead and set up our partial fraction decomposition.

00:52 So we'll have x cubed plus 3x squared plus x plus 9 over x squared plus 1 times x squared.

01:01 X squared plus 3 is equal to a x plus b over x squared plus 1 plus cx plus d over x squared plus 3 so we're going to need to multiply both sides by the denominator so we're going to get x cubed plus 3x squared plus x plus 9 is equal to so we get a x plus b times x squared plus 3 so what's that going to be and so we get a x squared plus 3ax plus bx squared plus 3b and then we get cx plus d times x squared plus 1 so that will be plus cx cubed plus cx plus d x squared plus d next we want to group by powers of x so we're going to get a plus c times x cubed and then what do we have for x squared so there's b and there's d and then what do we have for x so there's 3a plus c times x plus and then what do we have for constant so there's 3b plus d all right so that's equal to x cubed plus 3x squared plus x plus 9 okay so now we just just match basically.

02:42 So for x cubed we have a x plus so we have a plus c is equal to one for the x term for x squared so we get b plus d equals three and then we have three a plus c equals one and then what do we have so three b plus d is equal to nine right so if we subtract these two right we get negative three a is equal to zero that means a equal zero and c is an equal one if we do the same thing for these two we're gonna get negative three b is equal to negative six so that means b equals two and so if b equals 2, oh, that's not quite right.

03:58 So this should be negative 2.

04:03 And this should have been negative 2.

04:06 Either way, so a remains unchanged.

04:08 B is equal to 3.

04:10 So does that work? so if b is 3, then we get 3 plus d equals 3 and d equals 0...

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