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LearninDaMath
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∫[(2)/(x-3)(√(x+10))]dx U-Substitution → du/dx or dx/du, & why?
Okay, here I solve \int\frac{2}{(x+3)\sqrt{x+10}}dx in two ways. The problem I'm having however is during the U substitution. The first method, I take \frac{du}{dx}= (function in terms of x)
For the second method, I set substituted terms in terms of u and then take \frac{dx}{du}=function in terms of u)
So if I have indeed yielded the correct integral function using both methods, my question is is then based on my professor's suggestion that we use the second method for integrating various functions - particularly of this type with roots.
My question is, after choosing the terms to substitute for U, how do I look at a function to determine whether to set substituted terms into terms of x or u? In other words, how do I look at a function and determine whether to use the first method du/dx or second method dx/du?
Method 1:
\int\frac{2}{(x+3)\sqrt{x+10}}dxU substitution: Setting u equal to x terms, then du/dv
let u=\sqrt{x+10} then du/dv
\frac{du}{dx}=\frac{1}{2}(x+10)^{\frac{1}{2}-\frac{2}{2}}(x+10)'
du=\frac{1}{2\sqrt{x+10}}dx → dx=2\sqrt{x+10}duThen replace those substitutions into the integral and simplify what I can:\int\frac{(2)(2)\sqrt{x+10}du}{(x+3)\sqrt{x+10}} → 4\int\frac{du}{(x+3)}
Then realize I still have an x term and figure out a way to arrange the previous substitutions to eliminate the remaining x term
[u^{2}-10=x] → [u^{2}-10+3=x+3] → [u^{2}-7=x+3]
4\int\frac{1}{u^{2}-7}du ...And then partial fractions from here
4\int\frac{1}{u^{2}-7} = \frac{A}{u+\sqrt{7}}+\frac{B}{u-\sqrt{7}} becomes
4[\frac{1}{2\sqrt7}ln|\sqrt{x+10}+\sqrt{7}|-\frac{1}{2\sqrt7}ln|\sqrt{x+10}+\sqrt{7}|]+C
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___________________________________________________Method 2:
\int\frac{2}{(x+3)\sqrt{x+10}}dx
let u=\sqrt{x+10} set in terms of u
x=u^{2}-10
\frac{dx}{du}=2u
dx=2udu
\int\frac{(2)(2u)du}{u^{2}-10-3u}
4\int\frac{u}{(u-5)(u+2)}du And then partial fractions...
4\int\frac{A}{(u-5)}+\frac{B}{(u+2)}du
...Solving for A and B and integrating → 4[(\frac{5}{7}ln|\sqrt{x+10}-5|)+(\frac{2}{7}ln|\sqrt{x+10}+2|)]+C
Okay, here I solve \int\frac{2}{(x+3)\sqrt{x+10}}dx in two ways. The problem I'm having however is during the U substitution. The first method, I take \frac{du}{dx}= (function in terms of x)
For the second method, I set substituted terms in terms of u and then take \frac{dx}{du}=function in terms of u)
So if I have indeed yielded the correct integral function using both methods, my question is is then based on my professor's suggestion that we use the second method for integrating various functions - particularly of this type with roots.
My question is, after choosing the terms to substitute for U, how do I look at a function to determine whether to set substituted terms into terms of x or u? In other words, how do I look at a function and determine whether to use the first method du/dx or second method dx/du?
Method 1:
\int\frac{2}{(x+3)\sqrt{x+10}}dxU substitution: Setting u equal to x terms, then du/dv
let u=\sqrt{x+10} then du/dv
\frac{du}{dx}=\frac{1}{2}(x+10)^{\frac{1}{2}-\frac{2}{2}}(x+10)'
du=\frac{1}{2\sqrt{x+10}}dx → dx=2\sqrt{x+10}duThen replace those substitutions into the integral and simplify what I can:\int\frac{(2)(2)\sqrt{x+10}du}{(x+3)\sqrt{x+10}} → 4\int\frac{du}{(x+3)}
Then realize I still have an x term and figure out a way to arrange the previous substitutions to eliminate the remaining x term
[u^{2}-10=x] → [u^{2}-10+3=x+3] → [u^{2}-7=x+3]
4\int\frac{1}{u^{2}-7}du ...And then partial fractions from here
4\int\frac{1}{u^{2}-7} = \frac{A}{u+\sqrt{7}}+\frac{B}{u-\sqrt{7}} becomes
4[\frac{1}{2\sqrt7}ln|\sqrt{x+10}+\sqrt{7}|-\frac{1}{2\sqrt7}ln|\sqrt{x+10}+\sqrt{7}|]+C
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________Method 2:
\int\frac{2}{(x+3)\sqrt{x+10}}dx
let u=\sqrt{x+10} set in terms of u
x=u^{2}-10
\frac{dx}{du}=2u
dx=2udu
\int\frac{(2)(2u)du}{u^{2}-10-3u}
4\int\frac{u}{(u-5)(u+2)}du And then partial fractions...
4\int\frac{A}{(u-5)}+\frac{B}{(u+2)}du
...Solving for A and B and integrating → 4[(\frac{5}{7}ln|\sqrt{x+10}-5|)+(\frac{2}{7}ln|\sqrt{x+10}+2|)]+C
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