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

[LearninDaMath]

∫[(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

 

[tiny-tim]

Hi LearninDaMath!

I'm not sure I see any difference.

As a general principle, you should certainly get rid of all x's as soon as possible!

\int\frac{(2)(2u)du}{u^{2}-10-3u}

shouldn't this be \int\frac{(2)(2u)du}{(u^{2}-10+3)u} ?