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P x.ln((4-sqrt(x))/(4+sqrt(x)))
https://forumdematematica.org/viewtopic.php?f=6&t=65		 Página 1 de 1
Autor:  kinu [ 25 nov 2011, 05:00 ]
Título da Pergunta:  P x.ln((4-sqrt(x))/(4+sqrt(x)))
\(\displaystyle \int_{0}^{4}x.\ln\left(\frac{4-\sqrt{x}}{4+\sqrt{x}}\right)dx\)
Autor:  João P. Ferreira [ 25 nov 2011, 10:05 ]
Título da Pergunta:  Re: Integral
já experimentou fazer por partes... Derivar o ln e primitivar o x ?
Autor:  kinu [ 25 nov 2011, 10:35 ]
Título da Pergunta:  Re: Integral
i have tried like substitute \(\displaystyle \frac{4-\sqrt{x}}{4+\sqrt{x}}=t\Leftrightarrow \frac{8}{2\sqrt{x}}=-\left(\frac{t+1}{t-1}\right)\)

\(\displaystyle \frac{\sqrt{x}}{4} =\frac{1-t}{1+t}\)

but convert in very Lengthy Integrant

thank
Autor:  João P. Ferreira [ 25 nov 2011, 11:14 ]
Título da Pergunta:  Re: Integral
now based on your calculus you just have to calculate

\(\int 16(\frac{1-t}{1+t})^2 . ln(t).32.\frac{1-t}{1+t}.\frac{-2t}{(1+t)^2}dt\)

i suppose it gets too complicated :(

try with one of the trigonometric transformations
for example the tangent one

for sure it will work
Autor:  João P. Ferreira [ 25 nov 2011, 11:40 ]
Título da Pergunta:  Re: Integral
i think i found it :)

first you have to do it by parts

\(Pu'v=uv-Pv'u \ \ u'=x \ \ v=ln(\frac{4-\sqrt{x}}{4+\sqrt{x}})\)

then after derivating the ln and integrating the x you shall make a substitution like \(\sqrt{x}=t\)

then you'll have a rational function easy to integrate

let me know any results...
Autor:  João P. Ferreira [ 25 nov 2011, 19:30 ]
Título da Pergunta:  Re: Integral
We want to solve \(P x.ln(\frac{4-\sqrt{x}}{4+\sqrt{x}})\)

We shall use primitivation by parts

\(Pu'v=uv-Pv'u \\ \\ u'=x \ \ v=ln(\frac{4-\sqrt{x}}{4+\sqrt{x}})\)

\(u=\frac{x^2}{2}\)

\(v'=\frac{(\frac{4-\sqrt{x}}{4+\sqrt{x}})'}{\frac{4-\sqrt{x}}{4+\sqrt{x}}}=\frac{4}{\sqrt{x}(x-16)}\)

Then

\(P x.ln(\frac{4-\sqrt{x}}{4+\sqrt{x}}) = \frac{x^2}{2}.ln(\frac{4-\sqrt{x}}{4+\sqrt{x}})-P\frac{4}{\sqrt{x}(x-16)}.\frac{x^2}{2}\)

Now we just need to solve the primitive on the right side of the equation:

\(P\frac{2.x^2}{\sqrt{x}(x-16)}\)

To solve that, now we'll make a substitution \(\sqrt{x}=t \ <=> \ x=t^2 \ <=> \ x'=2t\)

Then we have to solve now

\(P\frac{2.t^4}{t(t^2-16)}.2t = 4P\frac{t^4}{t^2-16}\)

\(\frac{t^4}{t^2-16}\) it is a rational function that we'll need to divide and decomposite

it's possible to prove that

\(\frac{t^4}{t^2-16}=t^2+16+\frac{256}{t^2-16}\)

Now let's calculate

\(P\frac{t^4}{t^2-16}=P(t^2+16+\frac{256}{t^2-16})=\frac{t^3}{3}+16t+32.ln(\frac{4-t}{4+t})+C\)

Now we just need to combine everything together remembering that \(x=t^2\)

\(P x.ln(\frac{4-\sqrt{x}}{4+\sqrt{x}}) = \frac{x^2}{2}.ln(\frac{4-\sqrt{x}}{4+\sqrt{x}})-4(\frac{x^{3/2}}{3}+16\sqrt{x}+32.ln(\frac{4-\sqrt{x}}{4+\sqrt{x}}))+C\)

I suppose almost 100% sure that everything it is correct

Now you just need to aply the fundamental calculus theorem to solve the integral

You can confirm the result here

http://www.wolframalpha.com/input/?i=integrate+x*ln%28%284-sqrt%28x%29%29%2F%284%2Bsqrt%28x%29%29%29

Take care
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