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# ∫〖(7+x+3x^2)/(x+x^3 ) dx〗

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∫〖(7+x+3x^2)/(x+x^3 ) dx〗

May 21, 2015

#1
+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
+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.

Melody May 21, 2015