Partial Fractions

[JessicaMX]

UPDATE: Solved!


Hello. I have a improper fraction, but numerator grade = denominator grade.

This is my work in progress, but is wrong. Please, help me .

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Done!

 

[DrPhil]

Hello. I have a improper fraction, but numerator grade = denominator grade.

This is my work in progress, but is wrong. Please, help me .

You have to divide out the "whole number" part of the improper fraction. By doing long division, I get

\(\displaystyle \dfrac{x^2 + 1}{x^2 - x} = 1 + \dfrac{x + 1}{x^2 - x} \)

that is, 1 with a remainder of (x + 1).

Now you have the integral of an integer for the first term, and you can do your partial fractions on the second term.

 

[soroban]

Hello, JessicaMX!

Hello. I have a improper fraction: numerator grade = denominator grade.
. . So you must apply Long Division.

\(\displaystyle \displaystyle\int\frac{x^2+1}{x^2-x}\,dx\)

\(\displaystyle \dfrac{x^2 + 1}{x^2-x} \;=\;1 + \dfrac{x+1}{x^2-x}\)


\(\displaystyle \text{We will decompose the fraction:}\)

. . \(\displaystyle \dfrac{x+1}{x(x-1)} \;=\;\dfrac{A}{x} + \dfrac{B}{x-1}\)

\(\displaystyle \text{Then: }\:x+1 \;=\;A(x-1) + Bx\)

\(\displaystyle \begin{array}{ccccccc}\text{Let }x = 0: & 1 \:=\: A(\text{-}1) + B(0) & \Rightarrow & A \:=\:\text{-}1 \\ \text{Let }x = 1: & 2 \:=\: A(0) + B(1) & \Rightarrow & B \:=\:2 \end{array}\)

\(\displaystyle \text{Hence: }\:\dfrac{x+1}{x(x-1)} \;=\;\text{-}\dfrac{1}{x} + \dfrac{2}{x-1}\)


\(\displaystyle \displaystyle\text{Therefore: }\:\int\left(1 - \frac{1}{x} + \frac{2}{x-1}\right)\,dx\)


\(\displaystyle \text{Got it?}\)