+0

# ∫〖(7+x+3x^2)/(x+x^3 ) dx〗

0
403
1

∫〖(7+x+3x^2)/(x+x^3 ) dx〗

Guest May 21, 2015

#1
+91900
+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
Sort:

#1
+91900
+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

### 7 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details