Solving $\int{\frac{2x^{4}-5x^{3}+11x^{2}-2x-10}{x^{3}-3x^{2}+7x-5}dx}$

Question

I'm trying to solve: $$\int{\frac{2x^{4}-5x^{3}+11x^{2}-2x-10}{x^{3}-3x^{2}+7x-5}dx}$$

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I thought first step should have been polynomial long-division: $$\mspace{210mu} \; 2x \\ x^{3} -3x^{2}+7x-5 {\Big |}\mspace{-4.1mu}\overline{\mspace{3 mu}2x^{4} - 5x^{3} + 11x^2 -2x-10} \\ \mspace{120mu} \underline{-2x^{4} - 6x^{3} + 14x^2 -10x+0} \\ \mspace{173mu} 0+x^{3}-3x^{2}+8x-10$$

Hence, $$\int{\frac{2x^{4}-5x^{3}+11x^{2}-2x-10}{x^{3}-3x^{2}+7x-5}dx}=\int{2x+\frac{x^{3}-3x^{2}+8x-10}{x^{3}-3x^{2}+7x-5}dx}$$

Dividing again:

$$\mspace{210mu} \; 1 \\ x^{3} -3x^{2}+7x-5 {\Big |}\mspace{-4.1mu}\overline{\mspace{3 mu}x^{3}-3x^{2}+8x-10} \\ \mspace{120mu} \underline{-x^{3}-3x^{2}+7x-5} \\ \mspace{173mu} x-5$$

Hence,

$$\int{\frac{2x^{4}-5x^{3}+11x^{2}-2x-10}{x^{3}-3x^{2}+7x-5}dx}=\int{2x+1+\frac{x-5}{x^{3}-3x^{2}+7x-5}dx}$$

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$$\int{\frac{2x^{4}-5x^{3}+11x^{2}-2x-10}{x^{3}-3x^{2}+7x-5}dx}=\int{2x\,dx}+\int{1\,dx}+\int{\frac{x-5}{x^{3}-3x^{2}+7x-5}dx}$$

First two integrals are elementary, $\int{2x\,dx}=x^{2}+C$ and $\int{1\,dx}=x+C$.

Solving third integral:

$$\int{\frac{x-5}{x^{3}-3x^{2}+7x-5}dx}$$

Next, step is factorization of denominator, I found root using rational root theorem:

$$a_{0}=5, a_{n}=x^{3}$$

$$\frac{1, 5}{1}=\frac{1}{1}=1$$

Thus, we get $x-1$, dividing the rest polynomial:

$$\frac{x^{3}-3x^{2}+7x-5}{x-1}=x^{2}-2x+5$$

Hence,

$$\int{\frac{x-5}{x^{3}-3x^{2}+7x-5}dx}=\int{\frac{x-5}{(x-1)(x^{2}-2x+5)}dx}$$

Partial fraction decomposition:

$$\int{\frac{x-5}{(x-1)(x^{2}-2x+5)}dx}=\int{\frac{A}{x-1}dx}+\int{\frac{B}{x^{2}-2x+5}dx}$$

Multiplying by denominator:

$$x-5=A(x^{2}-2x+1)+B(x-1)$$

Here comes root substitution, root of $(x-1)=1$, but root of $x^{2}-2x+1=\frac{2+\sqrt{0}}{2}=1$.

Thus both of the polynomials have the same root, and we get:

$$-4=A(0)+B(0)$$

Thus, I can't find $A$ nor $B$.

Problem:

In the last equation, both of the roots were 1, and by substitution I got zero. Was factorization incorrect? or did I fail to find proper roots?

Thank you!

Answer 1

Hint. Check the factorization of the denominator: $$x^3-3x^2+7x-5=(x - 1)(x^2 - 2 x + 5).$$ Moreover, in the partial fraction decomposition, you should try $$\frac{x-5}{x^{3}-3x^{2}+7x-5}=\frac{A}{x-1}+\frac{Bx+C}{x^{2}-2x+5}$$ where the three constants $A$, $B$, and $C$ have to be determined.

Answer 2

You are on the right track but you have made a mistake in factoring the denominator of $$ \int{\frac{x-5}{x^{3}-3x^{2}+7x-5}dx}=\int{\frac{x-5}{(x-1)(x^{2}-2x+1)}dx}$$

You have to have a 5 in your second parentheses.

Please check it again and go from there.