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Find the following fro-x+42) dxf(91x+12) dx= (Use C as the arbitrary constant )...

Question

Find the following fro-x+42) dxf(91x+12) dx= (Use C as the arbitrary constant )

Find the following fro-x+42) dx f(91x+12) dx= (Use C as the arbitrary constant )



Answers

Determine the following: $$\int \frac{x}{c} d x \,(c \text { a constant } \neq 0)$$

Last one in this little block of problems where we're dealing with Constance and are integral to solve the integral from B to B squared D. X over X. I'm gonna do my trick where? All right, this is one over X. The X. We'll then integrate it. The integral of one over X. S. L. N absolute value of X. We're going from B to be squared. So this will be that natural log the absolute value of the square minus. Natural log, absolute value of B. Okay, we can combine these together. We know B squared is always positive so we can drop the absolutely on that one. I'm gonna assume B is greater than zero for this problem. Okay. That should give me B squared over and be inside this natural log which will simplify down to the natural log would be okay if I don't assume that B is greater than zero. If I take that assumption off, I'll finish with is keep them an absolute value around it inside the Okay, so one of those two ounces should be fine here

We want to find values of C that will make this integral. True. So let's first just go ahead and integrate this expression. And then once we finish integrating, we can just go ahead and solve that equation for seats. So remember the way we read this? Is this inside integral here we first integrated with respect to why assuming X is a constant. And then once we do that, we can enter it with respect. X. Let's go to do that. So we'll have negative one to see of and integrating that inside expression with respect to why would give us 1/2 x times y squared Plus why? And we evaluate from 0 to 2 and now we can evaluate at two and zero, and doing that would give us two X plus two and then when we plug in zero, that all just cancels out. And now we integrate this with respect to X, and that's going to give us it looks like to or just x squared, plus two x evaluate from negative one to see. So plug it and see we're gonna get C squared, plus to see and then plugging in one that's going to give us one minus two, and so that would be negative one times one. So that gives us C squared plus to see is equal to one plus what and now this is supposed to be equal to for plus foresee. So let's go ahead and move everything over to the left hand side and then solve this quadratic equation. So moving everything over to the right side we should get. Zero is equal to C squared minus two. See by this three, and we can go ahead and factor this to be see minus three and C plus one. So since this is equal to zero, that's going to imply that C is equal to negative one or C is equal to three.

We have the ability to go. Sorry, even Russian Ese. In addition, off affects the ex speakers condition off X bar X digs So this can be solved using you and the been super so in this integration we have you d b minus d upon the except for you. Indeed, it's not the DX good news. This is the form along while integrating such problems because there are two functions eggs and even about X So x is a rhythmic Johnson and Pete Borax is exponential Fenson. Soon we have to solve this by using I lead. This is where I stand for in verse instant florid meat for all that bright D for geometric and e for exponentially. So the I have this installation off ex parte tax DX peoples executed or X minus one into the end of our ex The ex This is equal to X for X minus X. So we have Steven equals one and Cedric was minus one. This is the answer to the given problem

Constant C one and C two such that big F of X is an anti drift of a derivative of little F. A bex. Okay, so if Big F is an anti derivative, then we will take its derivative and set it equal to little F of X. And then we'll be able to find out what the constants are. So here we go. Yeah, After using the product rule on the 1st 1. So it's the first times the derivative of the second plus the second times the derivative of the first. And then this one is -C to eat the -X. And that should equal to three X each the -X. All right, So we have minus C one X each of the minus X plus each the minus X times C one minus C two. He called a three X each of the minus six. So -11 equals three and see one minus c two equals zero. So C one is negative three. And so C two. C one is negative three, so C two is Also negative of three. So, Big F of X is negative three X. Each of the minus x minus three E to the minus X. And it is an anti derivative of little F of X.


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