Question

Evaluate the following integral (Calculus 2) Please show step by step explanation!

Answer 1

Answer:

$\displaystyle \int \dfrac{x^2-4}{x^3-2x^2+x}\:\:\text{d}x=-4 \ln |x|+5 \ln|x-1|+\dfrac{3}{x-1}+\text{C}$

Step-by-step explanation:

Fundamental Theorem of Calculus

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given integral:

$\displaystyle \int \dfrac{x^2-4}{x^3-2x^2+x}\:\:\text{d}x$

Factor the denominator:

$\begin{aligned}\implies x^3-2x^2+x & = x(x^2-2x+1)\\& = x(x^2-x-x+1)\\& = x(x(x-1)-1(x-1))\\ & = x((x-1)(x-1))\\& = x(x-1)^2\end{aligned}$

$\implies \displaystyle \int \dfrac{x^2-4}{x^3-2x^2+x}\:\:\text{d}x=\displaystyle \int \dfrac{x^2-4}{x(x-1)^2}\:\:\text{d}x$

Take partial fractions of the given fraction by writing out the fraction as an identity:

$\begin{aligned}\dfrac{x^2-4}{x(x-1)^2} & \equiv \dfrac{A}{x}+\dfrac{B}{(x-1)}+\dfrac{C}{(x-1)^2}\\\\ \implies \dfrac{x^2-4}{x(x-1)^2} & \equiv \dfrac{A(x-1)^2}{x(x-1)^2}+\dfrac{Bx(x-1)}{x(x-1)^2}+\dfrac{Cx}{x(x-1)^2}\\\\ \implies x^2-4 & \equiv A(x-1)^2+Bx(x-1)+Cx \end{aligned}$

Calculate the values of A and C using substitution:

$\textsf{when }x=0 \implies -4=A(1)+B(0)+C(0) \implies A=-4$

$\textsf{when }x=1 \implies -3=A(0)+B(0)+C(1) \implies C=-3$

Therefore:

$\begin{aligned}\implies x^2-4 & \equiv -4(x-1)^2+Bx(x-1)-3x\\& \equiv -4(x^2-2x+1)+B(x^2-x)-3x\\& \equiv -4x^2+8x-4+Bx^2-Bx-3x\\& \equiv (B-4)x^2+(5-B)x-4\\\end{aligned}$

Compare constants to find B:

$\implies 1=B-4 \implies B=5$

Substitute the found values of A, B and C:

$\implies \displaystyle \int \dfrac{x^2-4}{x^3-2x^2+x}\:\:\text{d}x=\displaystyle \int -\dfrac{4}{x}+\dfrac{5}{(x-1)}-\dfrac{3}{(x-1)^2}\:\:\text{d}x$

$\textsf{Apply exponent rule} \quad \dfrac{1}{a^n}=a^{-n}$

$\implies \displaystyle \int \dfrac{x^2-4}{x^3-2x^2+x}\:\:\text{d}x=\displaystyle \int -\dfrac{4}{x}+\dfrac{5}{(x-1)}-3(x-1)^{-2}}\:\:\text{d}x$

$\boxed{\begin{minipage}{5 cm}\underline{Terms multiplied by constants}\\\\ \displaystyle \int ax^n\:\text{d}x=a \int x^n \:\text{d}x \end{minipage}}$

$\boxed{\begin{minipage}{4 cm}\underline{Integrating} \dfrac{1}{x} \\\\ \displaystyle \int \dfrac{1}{x}\:\text{d}x=\ln |x|+\text{C} \end{minipage}}$

$\boxed{\begin{minipage}{4 cm}\underline{Integrating ax^n }\\\\ \displaystyle \int ax^n\:\text{d}x=\dfrac{ax^{n+1}}{n+1}+\text{C} \end{minipage}}$

$\begin{aligned}\implies \displaystyle \int \dfrac{x^2-4}{x^3-2x^2+x}\:\:\text{d}x & =\displaystyle \int -\dfrac{4}{x}+\dfrac{5}{(x-1)}-3(x-1)^{-2}}\:\:\text{d}x\\\\& = \displaystyle -4\int \dfrac{1}{x}\:\:\text{d}x+5\int \dfrac{1}{(x-1)}\:\:\text{d}x-3 \int (x-1)^{-2}}\:\:\text{d}x\\\\& = \displaystyle -4 \ln |x|+5 \ln|x-1|-3 \int (x-1)^{-2}}\:\:\text{d}x\end{aligned}$

Use Integration by Substitution:

$\textsf{Let }u=(x-1) \implies \dfrac{\text{d}u}{\text{d}x}=1 \implies \text{d}x=\text{d}u}$

Therefore:

$\implies \displaystyle -4 \ln |x|+5 \ln|x-1|-3 \int (x-1)^{-2}}\:\:\text{d}x$

$\implies \displaystyle -4 \ln |x|+5 \ln|x-1|-3 \int u^{-2}}\:\:\text{d}u$

$\implies -4 \ln |x|+5 \ln|x-1|-\dfrac{3}{-1}u^{-2+1}+\text{C}$

$\implies -4 \ln |x|+5 \ln|x-1|+3u^{-1}+\text{C}$

$\implies -4 \ln |x|+5 \ln|x-1|+\dfrac{3}{u}+\text{C}$

Substitute back in u = (x - 1):

$\implies -4 \ln |x|+5 \ln|x-1|+\dfrac{3}{x-1}+\text{C}$

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