Inegration
- Thread starter anandcu3
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You have written this:
\(\displaystyle \int\frac{(2x+5)(x+2)(x-1)}{x-1}\;dx\)
This simplifies trivially for any region not containing x = 1, to \(\displaystyle \int (2x+5)(x+2)\;dx\)
The real question is your intent. Methinks you have forgotten your Order of Operations. If you mean this
\(\displaystyle \int\frac{(2x+5)}{(x-1)(x+2)(x-1)}\;dx\)
You probably should write that. Parentheses help [(2x+5)/((x-1)(x+2)(x-1))].
If you mean the second version, you'll need to warm up your Partial Fraction Decomposition skills.
\(\displaystyle \int\frac{(2x+5)(x+2)(x-1)}{x-1}\;dx\)
This simplifies trivially for any region not containing x = 1, to \(\displaystyle \int (2x+5)(x+2)\;dx\)
The real question is your intent. Methinks you have forgotten your Order of Operations. If you mean this
\(\displaystyle \int\frac{(2x+5)}{(x-1)(x+2)(x-1)}\;dx\)
You probably should write that. Parentheses help [(2x+5)/((x-1)(x+2)(x-1))].
If you mean the second version, you'll need to warm up your Partial Fraction Decomposition skills.
Hello, anandcu3!
This requires integration-by-parts ... three times.
[If you know the "tabular method", it's faster.]
. . \(\displaystyle \begin{Bmatrix}u &=& x^3 && dv &=& e^{ax}dx \\ du &=& 3x^2\,dx && v &=& \frac{1}{a}e^{ax}\end{Bmatrix}\)
\(\displaystyle \displaystyle I \;=\;\tfrac{1}{a}x^3e^{ax} - \tfrac{3}{a}\int x^2e^{ax}\,dx\)
. . \(\displaystyle \begin{Bmatrix}u &=& x^2 && dv &=& e^{ax}dx \\ du &=& 2x\,dx && v &=& \frac{1}{a}e^{ax} \end{Bmatrix}\)
\(\displaystyle \displaystyle I \;=\;\tfrac{1}{a}x^3e^{ax} - \tfrac{3}{a}\left[\tfrac{1}{a}x^2e^{ax} - \tfrac{2}{a}\int xe^{ax}dx\right]\)
\(\displaystyle \displaystyle I\;=\;\tfrac{1}{a}x^3e^{ax} - \tfrac{3}{a^2}x^2e^{ax} + \tfrac{6}{a^2}\int x e^{ax}\,dx \)
. . \(\displaystyle \begin{Bmatrix}u &=& x && dv &=& e^{ax}\,dx \\ du &=& dx && v &=& \frac{1}{a}e^{ax} \end{Bmatrix}\)
\(\displaystyle \displaystyle I \;=\;\tfrac{1}{a}x^3e^{ax} - \tfrac{3}{a^2}x^2e^{ax} + \tfrac{6}{a^2}\left[\tfrac{1}{a}xe^{ax} - \tfrac{1}{a}\int e^{ax}\,dx\right]\)
\(\displaystyle \displaystyle I \;=\;\tfrac{1}{a}x^3e^{ax} - \tfrac{3}{a^2}x^2e^{ax} + \tfrac{6}a^3}xe^{ax} - \tfrac{6}{a^3}\int e^{ax}\,dx\)
\(\displaystyle \displaystyle I \;=\;\tfrac{1}{a}x^3e^{ax} - \tfrac{3}{a^2}x^2e^{ax} + \tfrac{6}{a^3}xe^{ax} - \tfrac{6}{a^4}e^{ax} + C \)
\(\displaystyle I \;=\;\frac{1}{a^4}e^{ax}\left(a^3x^3 - 3a^2x^2 + 6ax - 6\right) + C\)
This requires integration-by-parts ... three times.
[If you know the "tabular method", it's faster.]
. . \(\displaystyle \begin{Bmatrix}u &=& x^3 && dv &=& e^{ax}dx \\ du &=& 3x^2\,dx && v &=& \frac{1}{a}e^{ax}\end{Bmatrix}\)
\(\displaystyle \displaystyle I \;=\;\tfrac{1}{a}x^3e^{ax} - \tfrac{3}{a}\int x^2e^{ax}\,dx\)
. . \(\displaystyle \begin{Bmatrix}u &=& x^2 && dv &=& e^{ax}dx \\ du &=& 2x\,dx && v &=& \frac{1}{a}e^{ax} \end{Bmatrix}\)
\(\displaystyle \displaystyle I \;=\;\tfrac{1}{a}x^3e^{ax} - \tfrac{3}{a}\left[\tfrac{1}{a}x^2e^{ax} - \tfrac{2}{a}\int xe^{ax}dx\right]\)
\(\displaystyle \displaystyle I\;=\;\tfrac{1}{a}x^3e^{ax} - \tfrac{3}{a^2}x^2e^{ax} + \tfrac{6}{a^2}\int x e^{ax}\,dx \)
. . \(\displaystyle \begin{Bmatrix}u &=& x && dv &=& e^{ax}\,dx \\ du &=& dx && v &=& \frac{1}{a}e^{ax} \end{Bmatrix}\)
\(\displaystyle \displaystyle I \;=\;\tfrac{1}{a}x^3e^{ax} - \tfrac{3}{a^2}x^2e^{ax} + \tfrac{6}{a^2}\left[\tfrac{1}{a}xe^{ax} - \tfrac{1}{a}\int e^{ax}\,dx\right]\)
\(\displaystyle \displaystyle I \;=\;\tfrac{1}{a}x^3e^{ax} - \tfrac{3}{a^2}x^2e^{ax} + \tfrac{6}a^3}xe^{ax} - \tfrac{6}{a^3}\int e^{ax}\,dx\)
\(\displaystyle \displaystyle I \;=\;\tfrac{1}{a}x^3e^{ax} - \tfrac{3}{a^2}x^2e^{ax} + \tfrac{6}{a^3}xe^{ax} - \tfrac{6}{a^4}e^{ax} + C \)
\(\displaystyle I \;=\;\frac{1}{a^4}e^{ax}\left(a^3x^3 - 3a^2x^2 + 6ax - 6\right) + C\)
- Joined
- Apr 12, 2005
- Messages
- 11,339
What would be the point of finding the partial fractions AFTER finding the anti-derivative?
Do the partial fractions so that you CAN find the anti-derivative.
Let's see what you get.
That factor in the denominator you mistyped as (2x + 5).
It is supposed to be (x + 2).
anandu,
also, you must have grouping symbols around the numerator:
(2x + 5)/[(x + 2)(x - 1)^2] or
\(\displaystyle (2x + 5)/[(x + 2)(x - 1)^2]\)
D
Deleted member 4993
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This does not match your original problem:
Are you saying you do not know how to find the antiderivative of:
\(\displaystyle \int \frac{du}{u} \ = \ ??\)
and
\(\displaystyle \int \frac{dv}{v^2} \ = \ ??\)
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