(adsbygoogle = window.adsbygoogle || []).push({}); ∫[(2)/(x-3)(√(x+10))]dx U-Substitution → du/dx or dx/du, & why???

Okay, here I solve [itex]\int\frac{2}{(x+3)\sqrt{x+10}}dx[/itex] in two ways.

The problem I'm having however is during the U substitution. The first method, I take [itex]\frac{du}{dx}= [/itex] (function in terms of x)

For the second method, I set substituted terms in terms of u and then take [itex]\frac{dx}{du}= [/itex]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:

[itex]\int\frac{2}{(x+3)\sqrt{x+10}}dx[/itex]

U substitution: Setting u equal to x terms, then du/dv

let [itex]u=\sqrt{x+10}[/itex] then du/dv

[itex]\frac{du}{dx}=\frac{1}{2}(x+10)^{\frac{1}{2}-\frac{2}{2}}(x+10)'[/itex]

[itex]du=\frac{1}{2\sqrt{x+10}}dx[/itex] → [itex]dx=2\sqrt{x+10}du[/itex]

Then replace those substitutions into the integral and simplify what I can:

[itex]\int\frac{(2)(2)\sqrt{x+10}du}{(x+3)\sqrt{x+10}}[/itex] → [itex]4\int\frac{du}{(x+3)}[/itex]

Then realize I still have an x term and figure out a way to arrange the previous substitutions to eliminate the remaining x term

[itex][u^{2}-10=x][/itex] → [itex][u^{2}-10+3=x+3][/itex] → [itex][u^{2}-7=x+3][/itex]

[itex]4\int\frac{1}{u^{2}-7}du[/itex] .....And then partial fractions from here

[itex]4\int\frac{1}{u^{2}-7}[/itex] = [itex]\frac{A}{u+\sqrt{7}}+\frac{B}{u-\sqrt{7}}[/itex] becomes

[itex]4[\frac{1}{2\sqrt7}ln|\sqrt{x+10}+\sqrt{7}|-\frac{1}{2\sqrt7}ln|\sqrt{x+10}+\sqrt{7}|]+C[/itex]

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Method 2:

[itex]\int\frac{2}{(x+3)\sqrt{x+10}}dx[/itex]

let [itex]u=\sqrt{x+10}[/itex] set in terms of u

[itex]x=u^{2}-10[/itex]

[itex]\frac{dx}{du}=2u[/itex]

[itex]dx=2udu[/itex]

[itex]\int\frac{(2)(2u)du}{u^{2}-10-3u}[/itex]

[itex]4\int\frac{u}{(u-5)(u+2)}du[/itex] And then partial fractions...

[itex]4\int\frac{A}{(u-5)}+\frac{B}{(u+2)}du[/itex]

........Solving for A and B and integrating →

[itex]4[(\frac{5}{7}ln|\sqrt{x+10}-5|)+(\frac{2}{7}ln|\sqrt{x+10}+2|)]+C[/itex]

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# Homework Help: ∫[(2)/(x-3)(√(x+10))]dx U-Substitution → du/dx or dx/du, & why?

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