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Evaluate the following Integral using partial fractions...

Question

Evaluate the following Integral using partial fractions

Evaluate the following Integral using partial fractions



Answers

Use partial fractions to evaluate the given integral. $$ \int \frac{1}{x(2 x+3)} d x $$

Okay, so you're in the interval of X over X Money Street. I was excised. Do we actually need to become the closest into a party faction and take the integral of it? So we're gonna rewrite. This is one over X 23 experts to equal to a its mystery be experts to explain where else you didn't get one of people to a X minus 28 But the X minus baby get. So we have that X hate speech in the *** to a minus three feet. We have a plus B, and he could, you know, go to a minus three big. He's the one that's most replied. This way to get to a plus two b is equal to zero. So we get that combining these two again negative b is equal to one to be as equals. Negative one. And we have that if being equals negative one. Hey, no, we have a is equal to one. Okay, so our car should fraction becomes in a girl. What over? Yes, it is 1 to 1 over X minus three plus b, which was Thank you, ***. One over X minus two. The ex. Hey, you know best we get this Ellen of exploiting three minus 11 of X minus two. Let's see. Minus using our Ellen properties, we have Ellen of X minus three over X mice to plus C.

For this problem we are to evaluate the integral using partial fractions. Now we begin by finding the partial fraction decomposition for the expression one over expose one times expose to Times X-plus three. Now, since one over X plus one times X plus two times X plus three has distinct linear factors in the denominator. Then the partial fractions will have The nominators X-plus one express to and X plus three. Now, since the denominators are linear than their corresponding numerator are constants, let's say A, B and C. Now our next step will be to find these values for A. B and C. To do that. We will multiply this equation by the LCD that is express one times X plus two times X plus three. So from here we have one equal to eight times x plus two times x plus three plus B. Times express one times express three plus C times X plus one times X plus two. Next we will fix values for X to be substituted to this equation. Not that we can pick any X value, but to be able to solve this easily we will use X values for which the the nominators of the partial fractions are zero. So in here we want to let X plus one equals zero and we get X equals negative one. So the equation becomes one equal to a times negative one plus two times negative one plus three plus in here for terms with factors express one, they become zero. So the second term become zero as well as the third term. So and here we have one equal to to A. Which means that Is equal to 1/2 next. We want to let explodes to equal zero. So if exports to is equal to zero this means that tax is equal to -2. And our equation becomes one equal to eight times in here. Since we have experts to and exposed to zero, this becomes zero plus we have B times negative two plus one Times -2 Plus three. Plus the third term become zero because you have a factor of exports to their which is equal to zero. So this is zero. And and here we have one equal to negative B. Which means that B is equal to negative one. Now let's say you want to set X plus three equals zero. So when expressed there is equal to zero. This means that Act is equal to negative three and our equation becomes one equal to in here. Since the first term has A factor of explosive speed. This becomes zero since expose 30 Plus the second term also become zero. Since you have expressed through there plus see time's we have negative three plus one times negative three plus two. Simplifying this, we have one equal to to see which means that C. is equal to one half. Therefore the partial fraction the composition for one over x plus one Times X-plus two Times X was three is just one half over X plus one minus one over expose to Plus we have one half over X plus three. Now this can be written as one half times one over X plus one -1 over express to Plus 1/2 times one over X plus three. Now, if we take the integral of the situation both sides with respect to accident we have the integral of one over X plus one times X plus two times X plus three D X. This is equal to one half times the integral of one over. Express won the x minus the integral of one over Express to dx Plus 1/2 times the integral of one over X plus three D X. Now this is equal to one half L. An absolute value of exports, one minus. Ellen absolute value of exports to Plus we have 1/2 Ln absolute value of X plus three and then plus C.

They were the integral of the X over X Cube rights. Hey, based on the previous problem, you know that excuse minus the decomposing and he composes into depart. Show fraction negative one over X plus one over two X plus one plus one over its U Ex. Manage one. Okay. And now we can just integrate this. This is just you could have a villain. X plus 1/2 mile End of X plus one plus 1/2 and x minus one. Let's see.

For this problem we are to evaluate the given integral using partial fractions now and here we begin by finding the partial fraction decomposition for the expression x minus one over x times x squared plus one. Now X -1 over X times X squared plus one. We'll have partial fractions with denominators, X and x squared plus one. Now for the linear denominator the corresponding numerator is a constant, that's a. And for the quadratic the nominator, the numerator is linear. So let's call it Bx plus C. Next you want to get rid of the denominators by multiplying this equation by its LCD. That's X times X squared plus one. and from here we have X -1 equal to eight times x squared plus one plus bx plus c times x expanding this. We have X -1 equal to eight times x squared plus one. Plus we have B x squared plus c. X combining like terms you have X -1 equal to a plus B times x squared plus we have C. X. And then plus mm. Next we will compare the coefficients at the left side and at the right side to make the equations maintaining a B and C. Now, for the coefficient of x squared since you don't have X squared at the left side, then we have a plus B which is equal to zero. And then for the coefficient of x we have C for the right side this is equal to one coming from the left side. And then lastly for the constant term we have A which is equal to -1. Now if is equal to negative one, then we have negative one plus beach is equal to zero gives us B which is equal to one. Now that we have values for A B and C. Then the partial fraction decomposition for our given expression is -1 over X plus we have B which is one. So that's one times X or X Plus C which is once or plus one. That's divided by X squared plus one. Which we can read further into negative one over X plus X over X squared plus one plus one over X Grade Plus one. Now integrating both sides with respect to X. We have the integral of x minus one over X times X squared plus one dx. This is equal to the integral of this. And then dx And then integrating term by term we have The integral of negative one over XDX plus the integral of x over x squared plus one dx plus the integral of one over X squared plus one. The X. Now the first integral that's just negative L. And absolute value of X. And for the second one you have to apply substitution. You want to let U equal to X squared plus one. This will give us one half. Do you equal to X dx? So and here we have plus the integral of one half. Do you over you plus in here. That's just tangent inverse of x. And then integrating the middle term. We have negative L. And absolute value of X plus one half L. An absolute value of you, which is just expert, plus one, which is always positive, so the absolute value is not necessary and then plus tangent inverse of X and then plus C.


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