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Integral with partial frac.
- Thread starter hrr379
- Start date
D
Deleted member 4993
Guest
I know I am doing this wrong. If some one be so kind to point out where I am going wrong?
\(\displaystyle \dfrac{d}{dx}\left [ln|2x+1|\right ] \ = \ \dfrac{1}{2x+1} \ * \ \dfrac{d}{dx}\left [2x+1\right ] \ = \ ??? \)
Hello, hrr379!
Your partial fractions are incorrect . . .
\(\displaystyle \displaystyle\frac{2}{(2x+1)(x+1)} \;=\;\frac{A}{2x+1} + \frac{B}{x+1}\)
. . . . . .\(\displaystyle 2 \;=\;A(x+1) + B(2x+1)\)
\(\displaystyle x = \text{-}1\!:\quad \overbrace{2 \;=\;B(\text{-}1)}^{\text{Here!}} \quad\Rightarrow\quad B = \text{-}2\)
\(\displaystyle x = \text{-}\frac{1}{2}\!: \quad 2 \;=\;A(\frac{1}{2}) \quad\Rightarrow\quad A = 4\)
\(\displaystyle \displaystyle\text{Therefore: }\:\frac{2}{(2x+1)(x+1)} \;=\;\frac{4}{2x+1} - \frac{2}{x+1}\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Always check your partial fractions before proceeding . . .
\(\displaystyle \displaystyle \frac{4}{2x+1} - \frac{2}{x+1} \;=\; \frac{4(x+1) - 2(2x+1)}{(2x+1)(x+1)} \)
. . . . . . . . . . . . . .\(\displaystyle \displaystyle=\;\frac{4x+4 - 4x - 2}{(2x+1)(x+1)} \;=\;\frac{2}{(2x+1)(x+1)} \)
. . . . . See?
Your partial fractions are incorrect . . .
I know I am doing this wrong.
If someone be so kind to point out where I am going wrong?
. . \(\displaystyle \displaystyle\int\frac{2}{2x^2+3x+1}\,dx\)
\(\displaystyle \displaystyle\frac{2}{(2x+1)(x+1)} \;=\;\frac{A}{2x+1} + \frac{B}{x+1}\)
. . . . . .\(\displaystyle 2 \;=\;A(x+1) + B(2x+1)\)
\(\displaystyle x = \text{-}1\!:\quad \overbrace{2 \;=\;B(\text{-}1)}^{\text{Here!}} \quad\Rightarrow\quad B = \text{-}2\)
\(\displaystyle x = \text{-}\frac{1}{2}\!: \quad 2 \;=\;A(\frac{1}{2}) \quad\Rightarrow\quad A = 4\)
\(\displaystyle \displaystyle\text{Therefore: }\:\frac{2}{(2x+1)(x+1)} \;=\;\frac{4}{2x+1} - \frac{2}{x+1}\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Always check your partial fractions before proceeding . . .
\(\displaystyle \displaystyle \frac{4}{2x+1} - \frac{2}{x+1} \;=\; \frac{4(x+1) - 2(2x+1)}{(2x+1)(x+1)} \)
. . . . . . . . . . . . . .\(\displaystyle \displaystyle=\;\frac{4x+4 - 4x - 2}{(2x+1)(x+1)} \;=\;\frac{2}{(2x+1)(x+1)} \)
. . . . . See?