Gönderen Konu: Belirsiz İntegral  (Okunma sayısı 973 defa)

Çevrimdışı ArtOfMathSolving

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Belirsiz İntegral
« : Haziran 21, 2016, 06:21:06 ös »
$$ \int \frac{1}{x^6+1} dx$$
integralinin değerini bulunuz.
Sıradan bir matematikçi...

Çevrimdışı stuart clark

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« Yanıtla #1 : Haziran 23, 2016, 01:21:54 ös »
 $\displaystyle \int\frac{1}{1+x^6}dx  = \frac{1}{2}\int\frac{\left(1+x^4\right)+\left(1-x^4\right)}{1+x^6}dx$


$\displaystyle  = \frac{1}{2}\int\frac{1+x^4}{1+x^6}dx+\frac{1}{2}\int\frac{1-x^4}{1+x^6}dx$


Now we will take $\displaystyle I = \frac{1}{2}\int\frac{1+x^4}{1+x^6}dx$ and $\displaystyle J = \frac{1}{2}\int\frac{1-x^4}{1+x^6}dx$


So first we will calculate value of $I$


So $\displaystyle I = \frac{1}{2}\int\frac{1+x^4}{1+x^6}dx = \frac{1}{2}\int\frac{(x^2+1)^2-2x^2}{1+x^6}dx$


So $\displaystyle I = \frac{1}{2}\int\frac{(x^2+1)^2}{(1+x^2)\cdot (x^4-x^2+1)}dx - \int\frac{x^2}{1+(x^3)^2}dx$


So $\displaystyle = \frac{1}{2}\int\frac{x^2+1}{x^4-x^2+1}-\int\frac{x^2}{1+(x^3)^2}dx$


So $\displaystyle  = \frac{1}{2}\int \frac{\left(1+\frac{1}{x^2}\right)}{\left(x-\frac{1}{x}\right)^2+1^2}-\int \frac{x^2}{1+(x^3)^2}dx$


Now Let $\displaystyle \left(x-\frac{1}{x}\right) = t \Leftrightarrow \left(1+\frac{1}{x^2}\right)dx = dt$ and $x^3 = u\Leftrightarrow 3x^2dx = du\displaystyle \Leftrightarrow dx = \frac{1}{3}du$


So $\displaystyle I = \frac{1}{2}\cdot \tan^{-1}\left(x-\frac{1}{x}\right) - \frac{1}{3}\cdot \tan^{-1}\left(x^3\right)+C_{1}$


Similarly we will calculate for $\displaystyle J = \frac{1}{2}\int\frac{1-x^4}{1+x^6}dx$


So $\displaystyle J = \frac{1}{2}\int\frac{(1-x^2)\cdot (1+x^2)}{(1+x^2)\cdot (x^4-x^2+1)}dx = -\frac{1}{2}\int\frac{x^2-1}{x^4-x^2+1}dx$


$\displaystyle J = -\frac{1}{2}\int \frac{\left(1-\frac{1}{x^2}\right)}{\left(x+\frac{1}{x}\right)^2+\left(\sqrt{3}\right)^2}dx$


Now Now Let $\displaystyle \left(x+\frac{1}{x}\right) = v \Leftrightarrow \left(1-\frac{1}{x^2}\right)dx = dv$


So $\displaystyle J = -\frac{1}{2}\cdot \frac{1}{2\sqrt 3}\cdot \ln \left|\frac{\left(x+\frac{1}{x}\right)-\sqrt{3}}{\left(x+\frac{1}{x}\right)+\sqrt{3}}\right|+C_{2}$


So $\displaystyle \int \frac{1}{1+x^6}dx = \frac{1}{2}\cdot \tan^{-1}\left(x-\frac{1}{x}\right) - \frac{1}{3}\cdot \tan^{-1}\left(x^3\right) - \frac{1}{4\sqrt{3}}\cdot \ln \left|\frac{\left(x+\frac{1}{x}\right)-\sqrt{3}}{\left(x+\frac{1}{x}\right)+\sqrt{3}}\right|+\mathbb{C}$

Çevrimdışı ArtOfMathSolving

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« Yanıtla #2 : Haziran 23, 2016, 01:44:50 ös »
Congratz for nice solution. ;D
Sıradan bir matematikçi...

 


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