Evaluate the integral. ∫4x2+2x+3x2+xdx\int_{}^{}\frac{4x^2+2x+3}{x^2+x}dx

4x+3ln⁡∣x∣−5ln⁡∣x+1∣+C4x+3\ln\left|x\right|-5\ln\left|x+1\right|+C

Evaluate the integral. ∫ 4 x 2 + 2 x + 3 x 2 + x d x \int_{}^{}\frac{4x^2+2x+3}{x^2+x}dx

this problem, we're asked to evaluate the following integral. And we have the integral of four x squared plus two x plus three over x squared plus x. Now how do I go about solving this problem? Well, right off the bat, looking at this problem here, I can see we have a rational function. Now, the first thing I would think about is trying to do some kind of variable substitution to make this rational function a bit more simple, or something we can approach. Now if I try to do a variable substitution down here, I know that I need to take the derivative of this denominator with respect to x, which would give me two x plus one, and that's not going to cancel or make anything more simple than the numerator. Now same thing if I set this numerator equal to u. If I went ahead and do did a variable substitution there, we would take the derivative here, which I can see would give me a x plus two, which is not going to help me with anything in the denominator. So variable substitution is not gonna be the strategy that we use for solving this problem. So what do I need to try now? Well, a good thing would be to try some kind of partial fraction decomposition to make this more simple. So if we can break this into multiple simpler fractions, we can integrate each one individually. Now right off the bat, I noticed something. There is an equal power in the numerator, an equal degree as there is in the the denominator. There's an x squared and an x squared. That's the highest degree in both. So that means what we first want to do is perform polynomial long division to get this more simple. So I'm gonna go ahead and do that over here. So we're gonna have x squared plus x, and that's going to be divided into this top piece. Now the reason that I'm doing this here is trying to break this down into a more simple integral of simpler fractions. Now what we first need to do is think about how many times can x squared go into four x squared. Well, that's gonna be four times. Four times x squared would give us four x squared, and then we'd have four times x, which gives us four x. Now you need to go ahead and subtract these two. Notice here the four x squareds would cancel. And so what we would have is two x minus four x, which is negative two x, then we could go ahead and bring down that three. Now I noticed that x squared couldn't go into two x here, so what we end up with is four and then a remainder of negative two x plus three. So So what this is going to be is the same thing as the integral of four plus, and then it's going to be negative 2x plus three over x squared plus x integrated with respect to x. So now we have this broken up. Now at this point, I know how to find the integral of four, so we're good there. But how do I find the integral of negative 2x plus three over x squared plus x? Well, I can't really break this down any further just by using any more of what we've done with the polynomial division, but what I can do is try to perform partial fractions on this portion instead. Now I can see that the degree in the numerator is less than the degree in the denominator, so we are good to do partial fractions. Now what I'm going to do is pull this specific piece aside to do this. So I'm gonna move down here to make some space. So what we're gonna have is negative two x plus three over x squared plus x. Now my first step here is going to be to factor this denominator. So factoring this, we're going to end up with negative 2x plus three on the top, and I can pull an x out here, giving us x times x plus one. Notice we have now split this into distinct linear factors. So what I can go ahead and do is I can say that this will be split into some constant a over the distinct linear factor x, plus some constant b over the distinct linear factor x plus one. So now we've gone ahead and we've simplified this fraction, and what I can do from here is take both sides of this equation and multiply it by the common denominator. In this case, I can see the common denominator is x times x plus one, which I need to distribute into each one of these terms. Now first off, let's multiply things right here. Multiplying this rational function times the common denominator will get the x times x plus one to cancel entirely. So all we're going to have is negative 2x plus three. Now next, I'm going to multiply this term with the a. If I multiply this out, notice the x is going to cancel completely. So all it's going to leave me with is a times x plus one. Now next, I'll move to this term right here. Notice here that if I multiply these out, the x plus one is going to completely cancel, leaving me with just b times x. Now at this point, I need to find my constants, a and b. And the way that I can do that is by making some strategic substitutions for x. I could also try to get a system of equations, but I think the strategic substitutions are, in general, going to be more fast. So what I'm first going to do is let x equal negative one. I'm doing this because I know negative one plus one would give me zero, eliminating this factor entirely. So we would have negative two times negative one, replacing this x with negative one is equal to and then that's plus three is equal to a times negative one plus one plus b times negative one. Notice this whole thing, negative one plus one, is gonna set everything there equal to zero. So all we're going to have is negative two times negative one, which is two, plus three, which is five, is equal to negative b. And that means that b is equal to negative five, so we have now solved for our b. So going back up here, we can say it's this whole thing minus five over x plus one, meaning that our 'b' right here is also negative five. Now next I need to solve for this 'a'. Well to find a, I can recognize something. If I say that x is equal to zero, notice zero times five here is just going to give us zero. So doing this, I can see that we have negative two times zero plus three is equal to a times zero plus one minus five times zero. This whole thing is gonna go to zero, and this whole piece will also go to zero since we're multiplying by zero in both cases. So we'll have that three is equal to a times zero plus one, which is a, meaning that a is equal to three. So we can take this a here, and we can replace it with three. And now we've gone ahead and split this into partial fractions. So going back up to the integral, keep in mind this whole thing, this whole rational function was just dealing with this piece of the integral. So I can rewrite out this integral to be the integral of four plus, and then we have three over x, minus five over x plus one, all integrated with respect to x. So how do I solve this integral? Well, what I need to do is go through here, and I need to integrate each portion separately. That's the rule when it comes to dealing with these integrals. So integrating four with respect to x, that's just gonna give me 4x. Now what about integrating three over x? Well, this three is a constant, so we're just gonna have plus three times the integral of x in the denominator, one over x, which would be the natural log of x, and then don't forget these absolute value bars. We'll have minus, and then we have the integral of this. Five is a constant, comes outside the integral. The integral of one over x plus one is the natural log of x plus one, and then we have plus the constant of integration c, and that is the solution to the problem. So hope you found this video helpful, and let's move on and get some more practice.

Evaluate the integral. ∫4x2+2x+3x2+xdx\int_{}^{}\frac{$$4x^2+2x+3$$}{$$x^2+x$$}dx

4x+3ln⁡∣x∣−5ln⁡∣x+1∣+C4x+3\ln\left|x\right|-5\ln\left|x+1\right|+C

4x+2ln⁡∣x∣+3ln⁡∣x−1∣+C4x+2\ln\left|x\right|+3\ln\left|x-1\right|+C

4x+3ln⁡∣x∣+5ln⁡∣x+1∣+C4x+3\ln\left|x\right|+5\ln\left|x+1\right|+C

4x+2ln⁡∣x∣−3ln⁡∣x+1∣+C4x+2\ln\left|x\right|-3\ln\left|x+1\right|+C

12. Techniques of Integration

Partial Fractions

Calculus

12. Techniques of Integration

Partial Fractions

Struggling with Calculus?

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Multiple Choice

Evaluate the integral.
$\int_{}^{} \frac{4 x^{2} + 2 x + 3}{x^{2} + x} d x$

$4 x + 2 \ln ∣ x ∣ + 3 \ln ∣ x - 1 ∣ + C$

$4 x + 3 \ln ∣ x ∣ + 5 \ln ∣ x + 1 ∣ + C$

$4 x + 3 \ln ∣ x ∣ - 5 \ln ∣ x + 1 ∣ + C$

$4 x + 2 \ln ∣ x ∣ - 3 \ln ∣ x + 1 ∣ + C$

Verified step by step guidance

Rewrite the given integral: $ \int \frac{4x^2 + 2x + 3}{x^2 + x} \, dx $. Start by factoring the denominator $ x^2 + x $ as $ x(x + 1) $. This will help simplify the expression.

Perform polynomial long division since the degree of the numerator ($ 4x^2 $) is greater than the degree of the denominator ($ x^2 $). Divide $ 4x^2 + 2x + 3 $ by $ x^2 + x $ to obtain a quotient and a remainder.

Express the integral as the sum of the quotient and the remainder divided by the original denominator: $ \int \left( \text{quotient} + \frac{\text{remainder}}{x(x + 1)} \right) \, dx $. Simplify the remainder term if necessary.

Decompose the fraction $ \frac{\text{remainder}}{x(x + 1)} $ into partial fractions. Write it as $ \frac{A}{x} + \frac{B}{x + 1} $, and solve for $ A $ and $ B $ by equating coefficients.

Integrate each term separately: the quotient term integrates directly, and the partial fractions $ \frac{A}{x} $ and $ \frac{B}{x + 1} $ integrate to logarithmic functions. Combine all terms and include the constant of integration $ C $.

8:01m

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Evaluate the integral. ∫4x2+2x+3x2+xdx\(\int\)_{}^{}\(\frac{4x^2+2x+3}{x^2+x}\)dx

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Partial Fractions practice set

Evaluate the integral.
$\int_{}^{} \frac{4 x^{2} + 2 x + 3}{x^{2} + x} d x$