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Solve the given problems by integration.Find the area bounded by $$y=\frac{1}{x+1}, x=0, y=0$$ and $$x=2$$.

Calculus 1 / AB

Chapter 28

Methods of Integration

Section 2

The Basic Logarithmic Form

Integrals

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Okay, you're given this function and they want us to find the area founded by this function and the X axis between X equals zero and X equals one. And if we take a quick look at this between zero and one, we have the graph looking something like this. Oops, that's not zero. That's supposed people one. So let's fix that. But there's a basic idea of the graph. So we're trying to find the area of that region right there. And so that's just going to be the integral of this function from 0 to 1, taken with respect, X. But first thing we have to do is we need to break this down using our partial fraction decomposition. I'm not gonna rewrite the original fraction, but the form of the decomposition. We have a linear factor, so that will be some value a over the X plus one, and then this is a quadratic factor. So that form has an ex term and a constant term. So that would be p X plus C over the X squared plus one. So this original fraction equals that decomposition We multiply everything by the common denominator. And when we do we will have each denominator reduce and we take the numerator times The other part. So X squared minds are X minus X squared numerator of the original fraction is gonna equal eight times the X squared plus one So he x squared plus a and then plus the b x plus C times the X plus one which will give us be x squared plus B x plus c x plus c And now we can equate our terms and their coefficients. So the X squared terms we're gonna have a negative one equals a plus b the ex terms we're gonna have one equals be posse and then the constant term of zero on the left hand side Belief will are constant term here from the right hand side. Hey, plus c. So now whatever method you're gonna use solve this system of equations. I did it on my graphing calculator using the matrix and the Matrix operations, and I got that A equals negative one be a zero and C equals positive one. So, in order to find the area here under this curve, we can rewrite are integral as you Nicole from 01 a negative one over X plus one plus one over X squared, plus one times or DX. And then we can integrate each of these individually so we would have negative in a girl from 01 of one over x plus one d x. We can just like you recall the X plus one and T u equals v X so that gives us one over you, do you? So that's the negative natural of the absolute value of exports one and then our one over X squared plus one. Well, that's the form you squared, plus a squared. So we have the integral of one over x squared plus one d x letting you equal X letting a equal one. We end up with that in a girl few remember from some Merkel your problems. We had that form that's one over a 1/1, which is one times the inverse tangent are you, which is X divided by a X divide about they want, It's just X. And then this would get evaluated between our limits off zero and one. So we'll plug in one. So we have negative natural Aga one plus one is two plus the inverse tangent of one minus the negative natural log of zero plus one as one plus the inverse tangent of zero. Now that second set of parentheses theme algorithm of one is just zero, and the inverse tangent of zero is zero. So that's all zero in verse. Tension of one is pi over four, so we have as our area by over four, minus the natural algorithm off to

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