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I'm reading Differential and Integral Calculus by Piskunov, and have read up to integration using substitution. I've learned how to solve integrals of the form: $$\int{\dfrac{Ax+B}{ax^2+bx+c}}dx$$ The integrals of the above form are solved by completing the square in the denominator, and decomposing the fraction into a sum such that the first fraction of the sum is of the form: $$\int{\dfrac{2ax+b}{ax^2+bx+c}}dx$$ and the second fraction in the sum is of the form: $$k\int{\dfrac{1}{ax^2+bx+c}}dx$$ By substituion, the above two terms are readily integrable.

I tried applying similar techniques to the following integral: $$\int{\dfrac{6x^4-5x^3+4x^2}{2x^2-x+1}}dx$$ But have not been able to simplify them enough. The numerator is always two degrees higher than the denominator, and the $x^2$ term doesn't yield to the substitutions that worked in the simpler forms I wrote about above.

This is what I got after some simplification: $$\dfrac{1}{4}\int{\dfrac{\left(3t^2+t-2\right)^2+71\left(t+1\right)^2}{\left(t^2+7\right)}}dt$$

$t$ is $4x-1$.

How can I solve the integral given in the question using only integration by substitution? In particular, can't use integration by parts.

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    $\begingroup$ Perform the long division first to arrive to something you knwo. $\endgroup$ May 30 at 7:40
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Note that $$6x^4-5x^3+4x^2=(3x^2-x)(2x^2-x+1)+x$$ So $$\int \frac {6x^4-5x^3+4x^2}{2x^2-x+1} dx =\int (3x^2-x)dx+\int \frac {x}{2x^2-x+1} dx$$ I think you can do it now, using the techniques you mention in the question.

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