Audrina Jackson

2022-07-14

Solving $\int \left(\frac{5{x}^{2}+3x-2}{{x}^{3}+2{x}^{2}}\right)$ via partial fractions

So, first of all, we must factorize the denominator:

${x}^{3}+2{x}^{2}=(x+2)\cdot {x}^{2}$

Great. So now we write three fractions:

$\frac{A}{{x}^{2}}+\frac{B}{x}+\frac{C}{x+2}$

Eventually we conclude that

$A(x+2)+B(x+2)(x)+C({x}^{2})=5{x}^{2}+3x-2$

So now we look at what happens when x=−2:

C=12

When x=0:

A=−1And now we are missing B, but we can just pick an arbitrary number for x like... 1:

B=−1

So, first of all, we must factorize the denominator:

${x}^{3}+2{x}^{2}=(x+2)\cdot {x}^{2}$

Great. So now we write three fractions:

$\frac{A}{{x}^{2}}+\frac{B}{x}+\frac{C}{x+2}$

Eventually we conclude that

$A(x+2)+B(x+2)(x)+C({x}^{2})=5{x}^{2}+3x-2$

So now we look at what happens when x=−2:

C=12

When x=0:

A=−1And now we are missing B, but we can just pick an arbitrary number for x like... 1:

B=−1

Zachery Conway

Beginner2022-07-15Added 7 answers

You have an algebra error. When you put x=−2, the constraint equation collapses to 4C=12, so C=3.

Your method for determining B is correct – the constraining identity must hold for any x. Putting x=1 is a fine choice as the arithmetic is easier. However the incorrect value for C will affect the value you determine for B.

Your method for determining B is correct – the constraining identity must hold for any x. Putting x=1 is a fine choice as the arithmetic is easier. However the incorrect value for C will affect the value you determine for B.

Jamison Rios

Beginner2022-07-16Added 6 answers

I think this is the easiest way.

$(B+C){x}^{2}+(A+2B)x+2A=5{x}^{2}+3x-2$

$\begin{array}{rlrl}2A& =-2& \to A& =-1\\ A+2B& =-1+2B=3& \to B& =2\\ B+C& =2+C=5& \to C& =3\end{array}$

$(B+C){x}^{2}+(A+2B)x+2A=5{x}^{2}+3x-2$

$\begin{array}{rlrl}2A& =-2& \to A& =-1\\ A+2B& =-1+2B=3& \to B& =2\\ B+C& =2+C=5& \to C& =3\end{array}$

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