# Thread: Integration by splitting the numerator

1. ## Integration by splitting the numerator

Integral of (5-x^2)dx / sqrt(3-2x-x^2)

By splitting the numerator, thanks for the help!

2. ## Re: Integration by splitting the numerator

Originally Posted by pikachu975
Integral of (5-x^2)dx / sqrt(3-2x-x^2)

By splitting the numerator, thanks for the help!
$\noindent One way is as follows (general method when the numerator is a quadratic and denominator is square root of a quadratic). Say we want to find \int \frac{q(x)}{\sqrt{Q(x)}}\,\mathrm{d}x, where Q(x) is a degree 2 polynomial and q(x) is as well.$

$\noindent 1. Write q(x) = aQ(x) + bQ'(x) + c for some wise choice of constants a,b,c (again you can try and show as an exercise that such constants will always exist (and in fact will be unique)).$

$\noindent 2. Observe that this causes the integral to become$

$a\int \sqrt{Q(x)}\,\mathrm{d}x + b\int\frac{Q'(x)}{\sqrt{Q(x)}}\,\mathrm{d}x + c\int \frac{1}{\sqrt{Q(x)}}\,\mathrm{d}x.$

$\noindent \Big{(}For the first integral we used that \frac{Q(x)}{\sqrt{Q(x)}} = \sqrt{Q(x)}.\Big{)}$

$\noindent 3. Evaluate each integral. (Each of these integrals has a standard method of solving and you are expected to be able to do each of them.)$

3. ## Re: Integration by splitting the numerator

Originally Posted by InteGrand
$\noindent One way is as follows (general method when the numerator is a quadratic and denominator is square root of a quadratic). Say we want to find \int \frac{q(x)}{\sqrt{Q(x)}}\,\mathrm{d}x, where Q(x) is a degree 2 polynomial and q(x) is as well.$

$\noindent 1. Write q(x) = aQ(x) + bQ'(x) + c for some wise choice of constants a,b,c (again you can try and show as an exercise that such constants will always exist (and in fact will be unique)).$

$\noindent 2. Observe that this causes the integral to become$

$a\int \sqrt{Q(x)}\,\mathrm{d}x + b\int\frac{Q'(x)}{\sqrt{Q(x)}}\,\mathrm{d}x + c\int \frac{1}{\sqrt{Q(x)}}\,\mathrm{d}x.$

$\noindent \Big{(}For the first integral we used that \frac{Q(x)}{\sqrt{Q(x)}} = \sqrt{Q(x)}.\Big{)}$

$\noindent 3. Evaluate each integral. (Each of these integrals has a standard method of solving and you are expected to be able to do each of them.)$
Thanks again for the help, can you please show me how to write the (5-x^2) in the form you stated, because I don't quite understand it. Thanks!

4. ## Re: Integration by splitting the numerator

Originally Posted by pikachu975
Thanks again for the help, can you please show me how to write the (5-x^2) in the form you stated, because I don't quite understand it. Thanks!
$\noindent In your particular example it is q(x) = 5-x^{2} and Q(x) = 3-2x -x^{2}, so Q'(x) = -2-2x. So try finding constants a,b,c such that$

$5-x^{2}\equiv a\left(3-2x -x^{2}\right) + b(-2-2x) + c.$

$\noindent (You can find these similarly to how you find constants in partial fractions decomposition when it's written out like that. Or since this system is triangular'', you can find the constants easily by first equating x^{2} coefficients, then x, then constant terms. (Don't worry if you don't know what triangular means here.))$

5. ## Re: Integration by splitting the numerator

Originally Posted by InteGrand
$\noindent In your particular example it is q(x) = 5-x^{2} and Q(x) = 3-2x -x^{2}, so Q'(x) = -2-2x. So try finding constants a,b,c such that$

$5-x^{2}\equiv a\left(3-2x -x^{2}\right) + b(-2-2x) + c.$

$\noindent (You can find these similarly to how you find constants in partial fractions decomposition when it's written out like that. Or since this system is triangular'', you can find the constants easily by first equating x^{2} coefficients, then x, then constant terms. (Don't worry if you don't know what triangular means here.))$
I tried it and got a = 1, b = -1, c = 0, but I don't think it's right

5-x^2 = -ax^2 - 2x (a+b) + (3a-2b+c)
So a = 1 by equating coefficients
a+b = 0, so b = -1
3a-2b+c = 5
3 + 2 + c = 5
c = 0

Did I do something wrong?

6. ## Re: Integration by splitting the numerator

Originally Posted by pikachu975
I tried it and got a = 1, b = -1, c = 0, but I don't think it's right

5-x^2 = -ax^2 - 2x (a+b) + (3a-2b+c)
So a = 1 by equating coefficients
a+b = 0, so b = -1
3a-2b+c = 5
3 + 2 + c = 5
c = 0

Did I do something wrong?
$\noindent You did it right. You can (and should) check your answer, i.e. check that 5-x^{2} =\left(3-2x -x^{2}\right) -(-2-2x).$

7. ## Re: Integration by splitting the numerator

Hey, here's my solution. It took me a while tho...

$\int\frac{5-x^{2}}{\sqrt{3-2x-x^{2}}}dx\\=-\int\frac{x^2}{\sqrt{3-2x-x^{2}}}dx+5\int\frac{1}{\sqrt{3-2x-x^{2}}}dx\\now\hspace{0.1cm}3-2x-x^{2}=-(x^{2}+2x-3)\\=-((x+1)^{2}-4)\\=4-(x+1)^{2}\\I=-\int\frac{x^{2}}{\sqrt{4-(x+1)^{2}}}dx+5\int\frac{1}{\sqrt{4-(x+1)^{2}}}dx\\Let\hspace{0.1cm}J=-\int\frac{x^{2}}{\sqrt{4-(x+1)^{2}}}dx\\let\hspace{0.1cm}x+1=2sin\theta\\dx =2cos\theta d\theta\\J=-\int\frac{x^{2}cos\theta}{cos\theta}d\theta\\=-\int(2sin\theta-1)^{2}d\theta\\=-\int4sin^{2}\theta-4sin\theta cos\theta+1\hspace{0.1cm}d\theta\\=-\int3-2cos2\theta-2sin2\theta\hspace{0.1cm}d\theta\\=-[3\theta-sin2\theta+cos2\theta]+c\\=-3sin^{-1}(\frac{x+1}{2})+\frac{(x+1)\sqrt{3-2x-x^{2}}}{2}-1+\frac{(x+1)^{2}}{2}+c\\Let\hspace{0.1cm} K=5\int\frac{1}{\sqrt{4-(x+1)^{2}}}dx\\=5sin^{-1}(\frac{x+1}{2})+c\\I=J+K\\=2sin^{-1}(\frac{x+1}{2})+\frac{(x-3)\sqrt{3-2x-x^{2}}}{2}+c$

Note 10 attempts at editing it to get the equation to be valid... rip

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