# Hard rational Integral -Help

### Hard rational Integral -Help

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
Learningforcomp

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### Re: Hard rational Integral -Help

Math Tutor

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### Re: Hard rational Integral -Help

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

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

$$\frac{3x^2+ x+ 4}{x^4+ 3x^2+ 2}= \frac{Ax^3+ Bx^2+ 2Ax+ 2B+ Cx^3+ Dx^2+ Cx+ D}{x^4+ 3x^2+ 2}$$

$$\frac{3x^2+ x+ 4}{x^4+ 3x^2+ 2}= \frac{(A+ C)x^3+ (B+ D)x^2+ (2A+ C)x+ 2B+ D}{x^4+ 3x^2+ 2}$$

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.
Guest

### Re: Hard rational Integral -Help

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

The integral then is
$$\int \frac{x}{x^2+ 1}dx+ \int\frac{1}{x^2+ 1}dx- \int\frac{x}{x^2+ 2}dx+ \int\frac{2}{x^2+ 2}$$.

The two with "x" can be integrated with the substitutions $$u= x^2+ 1$$ and $$u= x^2+ 2$$. The other two are arctangent integrals.
Guest

### Re: Hard rational Integral -Help

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

Thank you very much!
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