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Need help with this difficult integral?
- Thread starter twohaha
- Start date
D
Deleted member 4993
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integrate: 1/(x-sqrt(2+x))
I tried changing it to
(x+sqrt(2+x))/(-2-x+x^2), separating the fraction, doing partial fraction for the first part and a trig sub for the second. No matter what i do, no trig sub really works...
Thanks!
using wolframalpha:
integral 1/(x-sqrt(2+x)) dx = 2/3 (2 log(2-sqrt(x+2))+log(sqrt(x+2)+1))+constant
Hello, twohaha!
\(\displaystyle \text{Let }u \,=\,\sqrt{x+2} \quad\Rightarrow\quad x \:=\:u^2-2 \quad\Rightarrow\quad dx \,=\,2u\,du\)
\(\displaystyle \displaystyle\text{Substitute: }\:\int \frac{2u\,du}{u^2-2 + u} \;=\;\int\frac{2u\,du}{u^2+u-2} \;=\;\int\frac{2u\,du}{(u-1)(u+2)}\)
\(\displaystyle \text{Partial Fractions: }\:\dfrac{2u}{(u-1)(u+2)} \:=\:\dfrac{A}{u-1} + \dfrac{B}{u+2}\)
. . . . . . . . . . . . . . . \(\displaystyle 2u \;=\;A(u+2) + B(u-1)\)
\(\displaystyle \begin{array}{cccccc}\text{Let }u=1: & 2 \:=\:A(3) + B(0) & \Rightarrow & A = \frac{2}{3} \\ \text{Let }u = \text{-}2: & \text{-}4 \:=\:A(0) + B(\text{-}3) & \Rightarrow & B = \frac{4}{3} \end{array}\)
\(\displaystyle \text{Hence: }\:\dfrac{2u}{(u-1)(u+2)} \;=\;\dfrac{\frac{2}{3}}{u-1} + \dfrac{\frac{4}{3}}{u+2}\)
\(\displaystyle \displaystyle\text{We have: }\:\tfrac{2}{3}\int\frac{du}{u-1} + \tfrac{4}{3}\int\frac{du}{u+2} \;=\;\tfrac{2}{3}\ln|u-1| + \tfrac{4}{3}\ln|u+2| + C \)
\(\displaystyle \text{Back-substitute: }\:\frac{2}{3}\ln\left|\sqrt{x+2} - 1\right| + \frac{4}{3}\ln\left|\sqrt{x+2}+2\right| + C\)
\(\displaystyle \displaystyle\text{Integrate: }\:\int \frac{dx}{x-\sqrt{x+2}}\)
\(\displaystyle \text{Let }u \,=\,\sqrt{x+2} \quad\Rightarrow\quad x \:=\:u^2-2 \quad\Rightarrow\quad dx \,=\,2u\,du\)
\(\displaystyle \displaystyle\text{Substitute: }\:\int \frac{2u\,du}{u^2-2 + u} \;=\;\int\frac{2u\,du}{u^2+u-2} \;=\;\int\frac{2u\,du}{(u-1)(u+2)}\)
\(\displaystyle \text{Partial Fractions: }\:\dfrac{2u}{(u-1)(u+2)} \:=\:\dfrac{A}{u-1} + \dfrac{B}{u+2}\)
. . . . . . . . . . . . . . . \(\displaystyle 2u \;=\;A(u+2) + B(u-1)\)
\(\displaystyle \begin{array}{cccccc}\text{Let }u=1: & 2 \:=\:A(3) + B(0) & \Rightarrow & A = \frac{2}{3} \\ \text{Let }u = \text{-}2: & \text{-}4 \:=\:A(0) + B(\text{-}3) & \Rightarrow & B = \frac{4}{3} \end{array}\)
\(\displaystyle \text{Hence: }\:\dfrac{2u}{(u-1)(u+2)} \;=\;\dfrac{\frac{2}{3}}{u-1} + \dfrac{\frac{4}{3}}{u+2}\)
\(\displaystyle \displaystyle\text{We have: }\:\tfrac{2}{3}\int\frac{du}{u-1} + \tfrac{4}{3}\int\frac{du}{u+2} \;=\;\tfrac{2}{3}\ln|u-1| + \tfrac{4}{3}\ln|u+2| + C \)
\(\displaystyle \text{Back-substitute: }\:\frac{2}{3}\ln\left|\sqrt{x+2} - 1\right| + \frac{4}{3}\ln\left|\sqrt{x+2}+2\right| + C\)
Hello, twohaha!
\(\displaystyle \text{Let }u \,=\,\sqrt{x+2} \quad\Rightarrow\quad x \:=\:u^2-2 \quad\Rightarrow\quad dx \,=\,2u\,du\)
\(\displaystyle \displaystyle\text{Substitute: }\:\int \frac{2u\,du}{u^2-2 + u} \;=\;\int\frac{2u\,du}{u^2+u-2} \;=\;\int\frac{2u\,du}{(u-1)(u+2)}\) ===> sign error (see below)
\(\displaystyle \text{Partial Fractions: }\:\dfrac{2u}{(u-1)(u+2)} \:=\:\dfrac{A}{u-1} + \dfrac{B}{u+2}\)
. . . . . . . . . . . . . . . \(\displaystyle 2u \;=\;A(u+2) + B(u-1)\)
\(\displaystyle \begin{array}{cccccc}\text{Let }u=1: & 2 \:=\:A(3) + B(0) & \Rightarrow & A = \frac{2}{3} \\ \text{Let }u = \text{-}2: & \text{-}4 \:=\:A(0) + B(\text{-}3) & \Rightarrow & B = \frac{4}{3} \end{array}\)
\(\displaystyle \text{Hence: }\:\dfrac{2u}{(u-1)(u+2)} \;=\;\dfrac{\frac{2}{3}}{u-1} + \dfrac{\frac{4}{3}}{u+2}\)
\(\displaystyle \displaystyle\text{We have: }\:\tfrac{2}{3}\int\frac{du}{u-1} + \tfrac{4}{3}\int\frac{du}{u+2} \;=\;\tfrac{2}{3}\ln|u-1| + \tfrac{4}{3}\ln|u+2| + C \)
\(\displaystyle \text{Back-substitute: }\:\frac{2}{3}\ln\left|\sqrt{x+2} - 1\right| + \frac{4}{3}\ln\left|\sqrt{x+2}+2\right| + C\)
Error correction: should be
\(\displaystyle \int\frac{2u\, du}{u^2-2-u}=\int\frac{2u\, du}{u^2-u-2}=\int\frac{2u\, du}{(u+1)(u-2)}\)
Then proceed in similar fashion as Soroban describes.