Integration of $ \int_1^2 \frac{3x^3 + 3x^2 - 5x + 4}{3x^3 - 2x^2 +3x -2} \mathrm dx$

Question

Evaluate $$\int_1^2 \frac{3x^3 + 3x^2 - 5x + 4}{3x^3 - 2x^2 +3x -2}\mathrm dx$$I believe I should use the partial fractions technique to evaluate this integral, but I am not getting anywhere when I try it.

Answer 1

Note that $$\frac{3x^3 + 3x^2 - 5x + 4}{3x^3 - 2x^2 +3x -2}=\frac{3x^3 - 2x^2 +3x -2+\color{red}{5x^2-8x+6}}{3x^3 - 2x^2 +3x -2}=1+\frac{5x^2-8x+6}{3x^3 - 2x^2 +3x -2}$$ and $$3x^3-2x^2+3x-2=(3x-2)(x^2+1)$$ so $$\int_1^2\frac{3x^3 + 3x^2 - 5x + 4}{3x^3 - 2x^2 +3x -2}\,dx=[x]_1^2+\int_1^2\frac{5x^2-8x+6}{(3x-2)(x^2+1)}\,dx$$ Now can you use partial fractions?

Answer 2

\begin{align} \frac {3x^3 + 3x^2 - 5x + 4}{3x^3 -2x^2 + 3x - 2} & = \frac {(3x^3 - 2x^2 + 3x - 2) + 5x^2 - 8x + 6}{3x^3 -2x^2 + 3x - 2}\\ \\ &\ = 1 + \frac{5x^2 - 8x + 6}{3x^3 -2x^2 + 3x - 2} \end{align}

Factor the denominator

$$3x^3 -2x^2 + 3x - 2 = (3x-2)(x^2 + 1)$$

$$\frac {5x^3 -8x^2 +6}{(3x-2)(x^2+1)} = \frac {A}{3x-2}+ \frac {Bx+C}{x^2 + 1}$$

The "obvious" way to do this..

$$5x^2 -8x +6 = Ax^2 + A + 3Bx^2 + (-2B+3C)x - 2C$$

Giving a system of equations:

\begin{align} A+3B &= 5\\ -2B+3C &= -8\\ A - 2C &= 6 \end{align}

Here is a trick which you may, or may not find easier.

$$\frac {5x^3 -8x^2 +6}{(3x-2)(x^2+1)} = \frac {A}{3x-2}+ \frac {Bx+C}{x^2 + 1}$$

Multiply through by $(3x-2)$

$$\frac {5x^3 -8x^2 +6}{(x^2+1)} = A+ \frac {Bx+C}{x^2 + 1}(3x-2)$$

and evaluate at $x = 2/3$

$$\frac{\frac{26}{9}}{\frac{13}{9}} = 2 = A \implies A = 2 $$

$$\frac {3x^3 + 3x^2 - 5x + 4}{3x^3 -2x^2 + 3x - 2} = 1 + \frac {2}{3x- 2} + \frac {x-2}{x^2+1}$$

Answer 3

Apologies if it's unclear - I'm not sure how to get the polynomial division in Mathjax, if someone could advise me on how to do this it would be appreciated.

From here I'll use Mathjax however: $$5x^2-8x+6=A(x^2+1)+B(3x-2)$$ Let $A=px+q, B=rx+s$. If $p>0$, this equation transforms into a cubic so $p=0$.

Hence $qx^2+q+(rx+s)(3x-2)=5x^2-8x+6$

Expanding and collecting like terms we get: $$q+3r=5$$ $$3s-2r=-8$$ $$q-2s=6$$ I instantly checked $q=2$ and $r=1$ (seemed like an obvious first choice given the first statement) and it works with $s=-2$.

Hence: $$\frac{5x^2-8x+6}{(x^2+1)(3x-2)}=1+\frac{2}{3x-2}+\frac{x-2}{x^2+1}$$

Integrate these seperately and sum the results: $$\int_1^2{1 dx}=\bigg[x\bigg]^2_1=1$$

$$\int_1^2{\frac{2}{3x-2}dx}=2\int_1^2{\frac{1}{3x-2}dx}=2\bigg[\frac 13\ln|3x-2|\bigg]^2_1=\frac 23\ln(4)=\frac 43\ln(2)$$

$$\int_1^2{\frac{x-2}{x^2+1}dx}=\int_1^2{\frac{x}{x^2+1}dx}-\int_1^2{\frac{2}{x^2+1}dx}$$ Note the first of those is of the form $\int{\frac{f'(x)}{2f(x)}dx}$, this integral therefore is $\bigg[\frac 12\ln|x^2+1|\bigg]^2_1=\frac 12(\ln5-\ln2)$

The second half requires $u$-substitution. Let $x=\tan(u)$, then $dx=\sec^2(u) du$. Our integral becomes:

$$2\int_{\tan^{-1}(1)}^{\tan^{-1}(2)}{\frac{1}{sec^2(u)}\cdot \sec^2(u) du}=2\bigg[1\bigg]^{\tan^{-1}(2)}_{\tan^{-1}(1)} \because 1+\tan^2(u)=\sec^2(u)$$ $$=2\tan^{-1}(2)-\frac \pi2\approx 0.6435 \text{(this is also $\tan^{-1}\bigg(\frac 34\bigg)$)}$$

Summing all we get:$$1+\frac 56 \ln(2)+ \frac 12 \ln(5)-2\tan^{-1}(2)+\frac \pi2$$