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

1. Oct 9, 2012

### 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:

$\int\frac{2}{(x+3)\sqrt{x+10}}dx$

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

let $u=\sqrt{x+10}$ then du/dv

$\frac{du}{dx}=\frac{1}{2}(x+10)^{\frac{1}{2}-\frac{2}{2}}(x+10)'$

$du=\frac{1}{2\sqrt{x+10}}dx$ → $dx=2\sqrt{x+10}du$

Then 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 →

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

Last edited: Oct 9, 2012
2. Oct 9, 2012

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