Integration by Partial Fractions

[Jason76]

I can understand most of this except for one part:

\(\displaystyle \int \dfrac{2 - x}{x^{2} + x}dx\)

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

Now we multiply by the common denominator.

\(\displaystyle 2 - x = A(x + 1) + B(x)\)

Separately setting \(\displaystyle x(x + 1) = 0\) We find x = 0 and -1

We substitute those values (respectively) into the A and B equation \(\displaystyle A(x + 1) + B(x)\) to get A = 2 and B = -3.

x = 0

\(\displaystyle 2 - (0) = A(0 + 1) + B(0)\) A = 2

x = -1

\(\displaystyle 2 - (-1) = A(-1 + 1) + B(-1)\) A = -3

\(\displaystyle \int \dfrac{2 - x}{x^{2} + x}dx = \int \dfrac{2}{x} - \dfrac{3}{x + 1} dx\)

Now, I don't understand this part: \(\displaystyle = 2 \int \dfrac{1}{x}dx - 3\int \dfrac{1}{x + 1}dx\) What's going on here? Where did the 2 and 3 come from?

\(\displaystyle = 2 \ln (x) - 3 \ln (x + 1) + C\) I understand how the integral of 1 over a math expression becomes a \(\displaystyle \ln\) expression

 

[lookagain]

I can understand most of this except for one part:

\(\displaystyle \int \dfrac{2 - x}{x^{2} + x}dx\)

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

\(\displaystyle A(x + 1) + B(x)\)

Separately setting \(\displaystyle x(x + 1) = 0\) We find x = 0 and -1

We substitute those values to get A = 2 and B = -3.

\(\displaystyle \int \dfrac{2 - x}{x^{2} + x}dx = \int \dfrac{2}{x} > > + < < \dfrac{3}{x + 1} dx\) No, it is supposed to be a minus sign.

Now, I don't understand this part: \(\displaystyle 2 \int \dfrac{1}{x}dx > > + < < 3\int \dfrac{1}{x + 1}dx\) See above. What's going on here? Where did the 2 and 3 come from?

\(\displaystyle 2 \ln (x) - 3 \ln > > (x + 1) < < + C\) No, there has to be absolute value bars around "x + 1." I understand how the integral of 1 over a math expression becomes a \(\displaystyle \ln\) expression

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[Jason76]

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The minus signs have been put in. Original post edited.

 

[DrPhil]

\(\displaystyle \displaystyle \int \dfrac{2 - x}{x^{2} + x}dx = \int \left( \dfrac{2}{x} - \dfrac{3}{x + 1} \right) dx\)

Now, I don't understand this part: \(\displaystyle \displaystyle \ \ \ = 2 \int \dfrac{1}{x}dx - 3\int \dfrac{1}{x + 1}dx\ \ \ \) What's going on here? Where did the 2 and 3 come from?

\(\displaystyle = 2 \ln (x) - 3 \ln (x + 1) + C\) I understand how the integral of 1 over a math expression becomes a \(\displaystyle \ln\) expression

The integral of a sum it the sum of the integrals. All that happened between those two lines is to write separate integrations for the two terms, and to bring the constants A=2 and B=-3 outside the integral signs. Is that clear now?