May 19th, 2017, 11:06 AM  #1 
Member Joined: Jan 2017 From: California Posts: 64 Thanks: 5  Stuck on this integral
Hi guys. What is the best way to integrate this? I can't seem to wrap my head around it. Last edited by skipjack; May 19th, 2017 at 11:48 PM. 
May 19th, 2017, 11:34 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,551 Thanks: 1260 
if the integrand is $\left(x + \dfrac{5}{9}\right)^{3/2}$, then the antiderivative is $\dfrac{2}{5}\left(x + \dfrac{5}{9}\right)^{5/2} + C$ if the integrand is $\left(\dfrac{x + 5}{9}\right)^{3/2} = \dfrac{(x+5)^{3/2}}{27}$, then the antiderivative is $\dfrac{2}{135}\left(x + 5 \right)^{5/2} + C$ 
May 19th, 2017, 11:53 AM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,784 Thanks: 2197 Math Focus: Mainly analysis and algebra 
It looks ideal for a simple $u$ substitution.

May 20th, 2017, 02:41 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 17,221 Thanks: 1294 
To try to find $\displaystyle \!\int\! f(x)dx$ using the substitution $x = g(u)$, where $g(u)$ is invertible and differentiable, find $\dfrac{dx}{du} = g^\prime\!(u)$, then use $\displaystyle \!\int\! f(x)dx = \!\int\! f(g(u))\frac{dx}{du}du = \!\int\! f(g(u))g^\prime\!(u)du$. After integrating, write the result in terms of $x$ (using the inverse of $g$ if necessary). Explicitly specifying the domain of $g$ is advisable to assist with that and simplifying $f(g(u))g^\prime\!(u)$. To find, for example, $\displaystyle \!\int\! \frac{1}{x^2\sqrt{1  x^2}} dx$, where $0 < x < 1$, let $x = \cos(u)$, where $0 < u < \pi/2$, so that $\dfrac{dx}{du} = \sin(u)$ and $\sqrt{1  x^2} = \sin(u)$. This gives $\displaystyle \!\int\! \frac{1}{\cos^2(u)\sin(u)} (\sin(u)) du = \!\int\! \sec^2(u)du = \tan(u) + \text{C} = \frac{\sin(u)}{\cos(u)} + \text{C}$ (where $\text{C}$ is a constant), which equals $\dfrac{\sqrt{1  x^2}}{x} + \text{C}$. 

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