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Question: How do you integrate \(\dfrac{{{x}^{4}}+{{x}^{3}}+{{x}^{2}}+1}{{{x}^{2}}+x-2}\) using partial fracti...

How do you integrate x4+x3+x2+1x2+x−2\dfrac{{{x}^{4}}+{{x}^{3}}+{{x}^{2}}+1}{{{x}^{2}}+x-2} using partial fractions?

Explanation

Solution

We first try to describe the requirement and the process of finding the partial fractions. Then we factorise the denominator of x4+x3+x2+1x2+x−2\dfrac{{{x}^{4}}+{{x}^{3}}+{{x}^{2}}+1}{{{x}^{2}}+x-2}. We form the numerator of 5x−1x2−x−2\dfrac{5x-1}{{{x}^{2}}-x-2} with the factors. We simplify the equation 5x−1x2−x−2=Ax−2+Bx+1\dfrac{5x-1}{{{x}^{2}}-x-2}=\dfrac{A}{x-2}+\dfrac{B}{x+1} to find the partial fraction form.

Complete step by step solution:
For our given problem x4+x3+x2+1x2+x−2\dfrac{{{x}^{4}}+{{x}^{3}}+{{x}^{2}}+1}{{{x}^{2}}+x-2}, we first break x4+x3+x2+1{{x}^{4}}+{{x}^{3}}+{{x}^{2}}+1 into factors.
We have x4+x3+x2+1=x2(x2+x−2)+3x2+1=x2(x2+x−2)+3(x2+x−2)−3x+7{{x}^{4}}+{{x}^{3}}+{{x}^{2}}+1={{x}^{2}}\left( {{x}^{2}}+x-2 \right)+3{{x}^{2}}+1={{x}^{2}}\left( {{x}^{2}}+x-2 \right)+3\left( {{x}^{2}}+x-2 \right)-3x+7.
x4+x3+x2+1x2+x−2 =x2(x2+x−2)+3(x2+x−2)−3x+7x2+x−2 =x2+3−3x−7x2+x−2 \begin{aligned} & \dfrac{{{x}^{4}}+{{x}^{3}}+{{x}^{2}}+1}{{{x}^{2}}+x-2} \\\ & =\dfrac{{{x}^{2}}\left( {{x}^{2}}+x-2 \right)+3\left( {{x}^{2}}+x-2 \right)-3x+7}{{{x}^{2}}+x-2} \\\ & ={{x}^{2}}+3-\dfrac{3x-7}{{{x}^{2}}+x-2} \\\ \end{aligned}
Now we apply partial fractions for 3x−7x2+x−2\dfrac{3x-7}{{{x}^{2}}+x-2}
Using middle-term factorisation we get
x2+x−2 =x2+2x−x−2 =x(x+2)−(x+2) =(x+2)(x−1) \begin{aligned} & {{x}^{2}}+x-2 \\\ & ={{x}^{2}}+2x-x-2 \\\ & =x\left( x+2 \right)-\left( x+2 \right) \\\ & =\left( x+2 \right)\left( x-1 \right) \\\ \end{aligned}
Now we have to arrange the numerator 3x−73x-7 in the binary addition and subtraction of the same factors of x2+x−2{{x}^{2}}+x-2 which are (x+2),(x−1)\left( x+2 \right),\left( x-1 \right).
We can do that directly or apply the form where 3x−7x2+x−2=Ax+2+Bx−1\dfrac{3x-7}{{{x}^{2}}+x-2}=\dfrac{A}{x+2}+\dfrac{B}{x-1}.
We simplify the Ax+2+Bx−1\dfrac{A}{x+2}+\dfrac{B}{x-1}. We get Ax+2+Bx−1=A(x−1)+B(x+2)(x+2)(x−1)\dfrac{A}{x+2}+\dfrac{B}{x-1}=\dfrac{A\left( x-1 \right)+B\left( x+2 \right)}{\left( x+2 \right)\left( x-1 \right)}.
So, 3x−7x2+x−2=Ax+2+Bx−1=A(x−1)+B(x+2)(x+2)(x−1)\dfrac{3x-7}{{{x}^{2}}+x-2}=\dfrac{A}{x+2}+\dfrac{B}{x-1}=\dfrac{A\left( x-1 \right)+B\left( x+2 \right)}{\left( x+2 \right)\left( x-1 \right)} gives 3x−7=A(x−1)+B(x+2)3x-7=A\left( x-1 \right)+B\left( x+2 \right).
Simplifying we get 3x−7=A(x−1)+B(x+2)=x(A+B)+(2B−A)3x-7=A\left( x-1 \right)+B\left( x+2 \right)=x\left( A+B \right)+\left( 2B-A \right).
Equating the variables and constants we get (A+B)=3;(2B−A)=−7\left( A+B \right)=3;\left( 2B-A \right)=-7.
Adding the equations, we get
(A+B)+(2B−A)=3−7 ⇒3B=−4 ⇒B=−43 \begin{aligned} & \left( A+B \right)+\left( 2B-A \right)=3-7 \\\ & \Rightarrow 3B=-4 \\\ & \Rightarrow B=\dfrac{-4}{3} \\\ \end{aligned}
This gives A=3−B=3−−43=133A=3-B=3-\dfrac{-4}{3}=\dfrac{13}{3}.
The partial fraction form becomes 3x−7x2+x−2=13/3x+2+−4/3x−1\dfrac{3x-7}{{{x}^{2}}+x-2}=\dfrac{{}^{13}/{}_{3}}{x+2}+\dfrac{{}^{-4}/{}_{3}}{x-1}.
The final form is x4+x3+x2+1x2+x−2=x2+3−13/3x+2+4/3x−1\dfrac{{{x}^{4}}+{{x}^{3}}+{{x}^{2}}+1}{{{x}^{2}}+x-2}={{x}^{2}}+3-\dfrac{{}^{13}/{}_{3}}{x+2}+\dfrac{{}^{4}/{}_{3}}{x-1}
Now we integrate ∫x4+x3+x2+1x2+x−2dx=∫(x2+3−13/3x+2+4/3x−1)dx\int{\dfrac{{{x}^{4}}+{{x}^{3}}+{{x}^{2}}+1}{{{x}^{2}}+x-2}dx}=\int{\left( {{x}^{2}}+3-\dfrac{{}^{13}/{}_{3}}{x+2}+\dfrac{{}^{4}/{}_{3}}{x-1} \right)dx}.

& \int{\left( {{x}^{2}}+3-\dfrac{{}^{13}/{}_{3}}{x+2}+\dfrac{{}^{4}/{}_{3}}{x-1} \right)dx} \\\ & =\dfrac{{{x}^{3}}}{3}+3x-\dfrac{13}{3}\log \left| x+2 \right|+\dfrac{4}{3}\log \left| x-1 \right|+c \\\ \end{aligned}$$ **The integral form of $\dfrac{{{x}^{4}}+{{x}^{3}}+{{x}^{2}}+1}{{{x}^{2}}+x-2}$ using partial fractions is $$\dfrac{{{x}^{3}}}{3}+3x-\dfrac{13}{3}\log \left| x+2 \right|+\dfrac{4}{3}\log \left| x-1 \right|+c$$.** **Note:** The form gets trickier when there are the same factors within its power form. We can simplify the form by taking the form of $Ax+B$ instead of A in case of quadratics. These partial fraction forms are required for special forms of integration.