Hard rational Integral -Help

Hard rational Integral -Help

Postby Learningforcomp » Fri Mar 26, 2021 1:30 pm

Hello I was given the following integral to solve
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But I dont know what to do It i thought of dividing it into partial fractions but I cant seem to be able to turn the denominator into one.
Could You help me please?
You time is deeply appreciated, Thank You
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Re: Hard rational Integral -Help

Postby Math Tutor » Fri Mar 26, 2021 2:27 pm


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Re: Hard rational Integral -Help

Postby Guest » Wed Apr 14, 2021 4:16 pm

The denominator is [tex]x^4+ 3x^2+ 2[/tex]. That is "quartic" but is of "quadratic form" since if we let [tex]y= x^2[/tex] we can write it [tex]y^2+ 3y+ 2= (y+ 1)(y+ 2)= (x^2+ 1)(x^2+ 2)[/tex]. Neither of those can be factored further (in real numbers) so the "partial fractions expansion' is
[tex]\frac{3x^2+ x+ 4}{x^4+ 3x^2+ 2}= \frac{Ax+ B}{x^2+ 1}+ \frac{Cx+ D}{x^2+ 2}[/tex].

One way to determine A, B, C, D is to add the fractions on the right side:
[tex]\frac{3x^2+ x+ 4}{x^4+ 3x^2+ 2}= \frac{(Ax+ B)(x^2+ 2)}{(x^2+ 1)(x^2+ 2)}+ \frac{(Cx+ D)(x^2+ 1)}{(x^2+ 1)(x^2+ 2)}[/tex]

[tex]\frac{3x^2+ x+ 4}{x^4+ 3x^2+ 2}= \frac{Ax^3+ Bx^2+ 2Ax+ 2B+ Cx^3+ Dx^2+ Cx+ D}{x^4+ 3x^2+ 2}[/tex]

[tex]\frac{3x^2+ x+ 4}{x^4+ 3x^2+ 2}= \frac{(A+ C)x^3+ (B+ D)x^2+ (2A+ C)x+ 2B+ D}{x^4+ 3x^2+ 2}[/tex]

Since this is to be true for all x, the coefficients of corresponding powers must be equal:
A+ C= 0
B+ D= 3
2A+ C= 1
2B+ D= 4

We have four linear equations to solve for A, B, C, and D.
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Re: Hard rational Integral -Help

Postby Guest » Thu Apr 15, 2021 4:33 pm

Since there has been no response for several days:
A+ C= 0
B+ D= 3
2A+ C= 1
2B+ D= 4

Subtracting A+ C= 0 from 2A+ C= 1, A= 1. Then A+ C= 1+ C= 0 so C= -1. Subtracting B+ D= 3 from 2B+ D= 4, B= 1, Then B+ D= 1+ D= 3 so D= 2.

$\frac{3x^2+ x+ 4}{x^4+ 3x^2+ 2}= \frac{x+ 1}{x^2+ 1}+ \frac{-x+ 2}{x^2+ 2}$
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Re: Hard rational Integral -Help

Postby Guest » Thu Apr 15, 2021 4:47 pm

The integral then is
[tex]\int \frac{x}{x^2+ 1}dx+ \int\frac{1}{x^2+ 1}dx- \int\frac{x}{x^2+ 2}dx+ \int\frac{2}{x^2+ 2}[/tex].

The two with "x" can be integrated with the substitutions [tex]u= x^2+ 1[/tex] and [tex]u= x^2+ 2[/tex]. The other two are arctangent integrals.
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Re: Hard rational Integral -Help

Postby Guest » Tue Sep 21, 2021 2:13 am

[tex]\int\frac{3x^{2}+x+4}{x^{4}+3x^{2}+2}dx=
\int\frac{3x^{2}+x+4}{x^{4}+2x^{2}+x^{2}+2}dx=
\int\frac{3x^{2}+x+4}{x^{2}(x^{2}+2)+1(x^{2}+2)}dx=
\int\frac{3x^{2}+x+4}{(x^{2}+1)(x^{2}+2)}dx=
\int\frac{(a_{1}x+a_{0})(x^{2}+2)+(b_{1}x+b_{0})(x^{2}+1)}{(x^{2}+1)(x^{2}+2)}dx=
\int\frac{(a_{1}x+a_{0})(x^{2}+2)}{(x^{2}+1)(x^{2}+2)}dx+\int\frac{(b_{1}x+b_{0})(x^{2}+1)}{(x^{2}+1)(x^{2}+2)}dx=
\int\frac{a_{1}x+a_{0}}{x^{2}+1}dx+\int\frac{b_{1}x+b_{0}}{x^{2}+2}dx=
\int\frac{a_{1}x}{x^{2}+1}dx+\int\frac{a_{0}}{x^{2}+1}dx+\int\frac{b_{1}x}{x^{2}+2}dx+\int\frac{b_{0}}{x^{2}+2}dx=
a_{1}\int\frac{x}{x^{2}+1}dx+a_{0}\int\frac{1}{x^{2}+1}dx+b_{1}\int\frac{x}{x^{2}+2}dx+b_{0}\int\frac{1}{x^{2}+2}dx=
a_{1}(\frac{1}{2}ln(x^{2}+1)+C_{1})+a_{0}(1arctan(1x)+C_{2})+b_{1}(\frac{1}{2}ln(x^{2}+2)+C_{3})+b_{0}(\frac{1}{ \sqrt{2}}arctan(\frac{1}{\sqrt{2}}x)+C_{4})=
1(\frac{1}{2}ln(x^{2}+1)+C_{1})+1(1arctan(1x)+C_{2})+(-1)(\frac{1}{2}ln(x^{2}+2)+C_{3})+2(\frac{1}{ \sqrt{2}}arctan(\frac{1}{\sqrt{2}}x)+C_{4})=
\frac{1}{2}ln(x^{2}+1)+C_{1}+1arctan(1x)+C_{2}+(-\frac{1}{2})ln(x^{2}+2)+(-C_{3})+\sqrt{2}arctan(\frac{1}{\sqrt{2}}x)+2C_{4}=
\frac{1}{2}ln(x^{2}+1)+C_{1}+arctanx+C_{2}-\frac{1}{2}ln(x^{2}+2)-C_{3}+\sqrt{2}arctan\frac{x}{\sqrt{2}}+2C_{4}=
(\frac{1}{2}ln(x^{2}+1)+arctanx-\frac{1}{2}ln(x^{2}+2)+\sqrt{2}arctan\frac{x}{\sqrt{2}})+(C_{1}+C_{2}-C_{3}+2C_{4})=
\frac{1}{2}ln(x^{2}+1)+arctanx-\frac{1}{2}ln(x^{2}+2)+\sqrt{2}arctan\frac{x}{\sqrt{2}}+C[/tex]
[tex]u_{1}=x^{2}+1[/tex]
[tex]du_{1}=2xdx[/tex]
[tex]\int\frac{x}{x^{2}+1}dx=\int\frac{1}{2u_{1}}du_{1}=\frac{1}{2}\int\frac{1}{u_{1}}du_{1}=\frac{1}{2}lnu_{1}+C_{1}=\frac{1}{2}ln(x^{2}+1)+C_{1}[/tex]
[tex]u_{2}=1x[/tex]
[tex]du_{2}=1dx[/tex]
[tex]\int\frac{1}{x^{2}+1}dx=1\int\frac{1}{u_{2}^{2}+1}du_{2}=1arctanu_{2}+C_{2}=1arctan(1x)+C_{2}[/tex]
[tex]u_{3}=x^{2}+2[/tex]
[tex]du_{3}=2xdx[/tex]
[tex]\int\frac{x}{x^{2}+2}dx=\int\frac{1}{2u_{3}}du_{1}=\frac{1}{2}\int\frac{1}{u_{3}}du_{1}=\frac{1}{2}lnu_{3}+C_{3}=\frac{1}{2}ln(x^{2}+2)+C_{3}[/tex]
[tex]u_{4}=\frac{1}{\sqrt{2}}x[/tex]
[tex]du_{4}=\frac{1}{\sqrt{2}}dx[/tex]
[tex]\int\frac{1}{x^{2}+2}dx=\frac{1}{\sqrt{2}}\int\frac{1}{u_{4}^{2}+1}du_{4}=\frac{1}{\sqrt{2}}arctanu_{4}+C_{4}=\frac{1}{\sqrt{2}}arctan(\frac{1}{\sqrt{2}}x)+C_{4}[/tex]
[tex](a_{1}x+a_{0})(x^{2}+2)+(b_{1}x+b_{0})(x^{2}+1)=
a_{1}x^{3}+a_{0}x^{2}+2a_{1}x+2a_{0}+b_{1}x^{3}+b_{0}x^{2}+b_{1}x+b_{0}=
(a_{1}+b_{1})x^{3}+(a_{0}+b_{0})x^{2}+(2a_{1}+b_{1})x^{1}+(2a_{0}+b_{0})x^{0}=
3x^{2}+x+4=
0x^{3}+3x^{2}+1x^{1}+4x^{0}[/tex]
[tex]a_{1}+b_{1}=0[/tex]
[tex]2a_{1}+b_{1}=1[/tex]
[tex]a_{0}+b_{0}=3[/tex]
[tex]2a_{0}+b_{0}=4[/tex]
[tex]a_{1}=0-b_{1}[/tex]
[tex]2(0-b_{1})+b_{1}=0-2b_{1}+b_{1}=0-b_{1}=1[/tex]
[tex]a_{0}=3-b_{0}[/tex]
[tex]2(3-b_{0})+b_{0}=6-2b_{0}+b_{0}=6-b_{0}=4[/tex]
[tex]b_{1}=0-1=-1[/tex]
[tex]a_{1}=0-(-1)=1[/tex]
[tex]b_{0}=6-4=2[/tex]
[tex]a_{0}=3-2=1[/tex]
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Re: Hard rational Integral -Help

Postby Guest » Tue Sep 21, 2021 9:54 am

Thank you very much!
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