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∫〖(7+x+3x^2)/(x+x^3 ) dx〗

 May 21, 2015

Best Answer 

 #1
avatar+99361 
+10

I just lost a huge page of LaTex.    

Anyway I'll give you the abridged version.

I will expand on it if you need me too.      

I split the numerator up to give me 3 seperate fractions to integrate.

 

$$\\\int \frac{x}{x+x^3}=\int \frac{1}{1+x^2}=atan(x)\\\\
\frac{7}{x+x^3} $ can be broken up using partial fractions into$\\\\\
\frac{7}{x}+\frac{-7x}{1+x^2}\qquad $ I can expand on this if you need me too$\\\\
$so our integral becomes$\\\\
\int\;\frac{7}{x}+\frac{-7x}{1+x^2}+\frac{1}{1+x^2}+\frac{3x}{1+x^2}\;dx\\\\
=\int\;\frac{7}{x}+\frac{-4x}{1+x^2}+\frac{1}{1+x^2}\;dx\\\\
=7ln(x)-2ln(1+x^2)+tan^{-1}(x)+c$$

 

I think that is correct.  

 May 21, 2015
 #1
avatar+99361 
+10
Best Answer

I just lost a huge page of LaTex.    

Anyway I'll give you the abridged version.

I will expand on it if you need me too.      

I split the numerator up to give me 3 seperate fractions to integrate.

 

$$\\\int \frac{x}{x+x^3}=\int \frac{1}{1+x^2}=atan(x)\\\\
\frac{7}{x+x^3} $ can be broken up using partial fractions into$\\\\\
\frac{7}{x}+\frac{-7x}{1+x^2}\qquad $ I can expand on this if you need me too$\\\\
$so our integral becomes$\\\\
\int\;\frac{7}{x}+\frac{-7x}{1+x^2}+\frac{1}{1+x^2}+\frac{3x}{1+x^2}\;dx\\\\
=\int\;\frac{7}{x}+\frac{-4x}{1+x^2}+\frac{1}{1+x^2}\;dx\\\\
=7ln(x)-2ln(1+x^2)+tan^{-1}(x)+c$$

 

I think that is correct.  

Melody May 21, 2015

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