#1**+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.

Melody
May 21, 2015

#1**+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