In Exercises 9–16, express the integrand as a sum of partial f...

Question

In Exercises 9–16, express the integrand as a sum of partial fractions and evaluate the integrals.

∫ (x + 3) / (2x³ - 8x) dx

Accepted Answer

Welcome back everyone. Evaluate the the integral 3 x minus 2 divided by x cubed plus x2 dx. For this problem we're going to use partial fraction decomposition. Let's rewrite our function 3 x minus 2 divided by. And now let's factor out the denominator that will be x 2 in parent x + 1. And we can now rewrite this expression as a divided by x plus b divided by x 2, right? Notice that we have x 2, which means that we have to include both x and x2 for the first two fractions. And for the third one, we're going to include our final linear factor x + 1. And now what we can do is basically find the common denominator, so that would be a multiplied by x. Squad, right, but essentially what we want to do is multiply by x only since essentially that will give us x 2 in the denominator followed by x + 1. So we get a x multiplied by x + 1. Plus b multiplied by x + 1 Plusc multiplied by x 2 and this must be equal to the original numerator 3x minus 2. And now let's go ahead and find the value of each. Constant A, B, and C. First of all, we can set x equals 0. Because this is the term. Adjacent to a and c, specifically, we have x, right? So if we plug in x equals 0, it will cancel out a and c, and it will leave us with b multiplied by 0 plus 1, which is b multiplied by. One. And this is going to be equal to 3 multiplied by 0 minus 2, which is -2. So now B is equal to -2. 2. We can now set x equals -1 because it will cancel out a and b, right, and it will leave us with c multiplied by -12 equals 3 multiplied by -1 is -3 minus 2, that's -5. So we get C equals -5, right? Number 3, since we already have B and C, we want to solve for A, right? To find a, we can compare the coefficients of x 2. In particular, we can show that 0 equals. A plus C. And therefore, A is going to be equal to c. Therefore, A is going to be negative of -5, which is 5. Well done And now let's go ahead and rewrite our integral as follows. That's integral of a divided by x. So that's 5 divided by x. Plus B divided by X2, so that's -2 divided by X2. Plus c divided by x + 1, which is -5 divided by x + 1 followed by d x. And now we can integrate each separately, right? Let's remember that we can factor out each constant, so that'd be 5. Integral of d x divided by x minus 2 integral dx divided by x 2. For this specific part, we can just rewrite it as x to the power of -2 d x. And then we're subtracting 5 integral dx divided by x plus 1. And then we can integrate because we have elementary functions for the first part we get 5. Ln of the absolute value of x, that's the integral of d x divided by x minus 2 multiplied by the integral of x, the power of -2 can be found using the power rule. So that'd be x, the power of -2 + 1, that's -1 divided by -1. Plus Actually minus because we have -5 in front. So we get -5 ln of the absolute value of x + 1. Plessy constant of integration see because we have An indefinite integral. And now we can simplify using the properties of logarithms. Let's go ahead and do that. We're going to have 5. Ln of the absolute value of x. Divided by x + 1. Notice that first of all we can factor out 5 and then we have the difference of 2 logs. So we basically combine them within a ratio, right under the same logarithm. And then we are adding. 2 x to the power of -1, which can be written as 2 divided by x plus a constant of integration c at the end. That's our final answer and thank you for watching.

In Exercises 9–16, express the integrand as a sum of partial fractions and evaluate the integrals.∫ (x + 3) / (2x³ - 8x) dx

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In Exercises 9–16, express the integrand as a sum of partial fractions and evaluate the integrals.
∫ (x + 3) / (2x³ - 8x) dx