Question:
Find the value of $\displaystyle \int \left(\frac{3x^2-9x+1}{3x^3-3x+3}\right)^2dx$.
Attempt:
$\displaystyle I = \int\left(\frac{x^2-3x+3^{-1}}{x^3-x+1}\right)^2dx$
Want to be able to go further, someone help me, thanks.
Find the value of $\int \left(\frac{3x^2-9x+1}{3x^3-3x+3}\right)^2dx$
0
$\begingroup$
real-analysis
integration
indefinite-integrals
factoring
fractions
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0now you must use a parfrac decomposition of the denominator. The result looks ugly – 2017-01-08
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1the result should be $$\frac{-9 x^2+27 x-26}{9 \left(x^3-x+1\right)}$$ – 2017-01-08
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0This calculator suggested to Apply Ostrogradsky's method: https://goo.gl/EwZ4rG – 2017-01-08
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0The value of an indefinite integral? Are you looking for a primitive or a value (like $\int_{- \infty}^{+ \infty} f(x) dx$)? – 2017-01-08
3 Answers
1
$\int\frac{(3x^2-9x+1)^2}{(3x^3-3x+3)^2}dx=-\frac{1}{9}\frac{9x^2-27x+26}{x^3-x+1}+C$
Because $(18x-27)(x^3-x+1)-(9x^2-27x+26)(3x^2-1)=-(3x^2-9x+1)^2$.
2
We can simplify $$I =\int (\frac {3x^2-9x+1}{3x^3-3x+3})^2 dx = \frac {1}{9} \int \frac {(3x^2-9x+1)^2}{(x^3-x+1)^2} dx = \frac {1}{9} I_1$$
To calculate $I_1$, we use the Ostrogadsky method. For the procedure to be used, see here.
For our integral we have, $$P (x)= (3x^2-9x+1)^2 $$ $$Q (x)=(x^3-x+1)^2$$ $$Q'(x)=2 (3x^2-1)(x^3-x+1) $$ $$Q_1(x)=Q_2 (x)=x^3-x+1$$ We then get the polynomials $$P_1 (x)= -9x^2+27x-26 \text { and } P_2 (x)= 0$$
Thus, $$I_1 =-\frac{9x^2-27x+26}{x^3-x+1} $$ Hope it helps.
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0Its x^3 below can you explain first step – 2017-01-08
2
$$\dfrac{d\left(\dfrac{Ax^2+Bx+C}{3x^3-3x+3}\right)}{dx}=\dfrac{(2Ax+B)(3x^3-3x+3)-(Ax^2+Bx+C)(9x^2-3)}{(3x^3-3x+3)^2}$$
$(3x^2-9x+1)^2=(2Ax+B)(3x^3-3x+3)-(Ax^2+Bx+C)(9x^2-3)$
$\iff9x^4-54x^3+90x^2-18x+1=(6A-9A)x^4+x^3(3B-6A-9B)+x^2(\cdots)+x(\cdots)+3B+3C$
Compare the constants and the coefficients of $x^4,x^3$ to find $A,B,C$