Integral of fraction - is this correct?

[username12345]

I am trying to find the integral of the following:

\int\left({\frac{3 + 5x - 6x^2 - 7x^3}{2x^2}}\right)dx

What I did was to split up the fraction like so:

\int\left({\frac{3}{2x^2}}\right)dx + \int\left({\frac{5x}{2x^2}}\right)dx + \int\left({\frac{6x^2}{2x^2}}\right)dx + \int\left({\frac{7x^2}{2x^2}}\right)dx

Simplified the fractions, worked out each integral then added them to get:

-\frac{3}{2x} + \frac{5}{2}\ln x - 3x - \frac{7}{4}x^2 + c

The text I am using has no answers and when I tried to use the integral calculator at http://www.numberempire.com/integralcalculator.php I get an answer that is a fraction.

Am I correct and is the process I used to solve this correct?

Thanks.

 

[tiny-tim]

I am trying to find the integral of the following:

\int({\frac{3 + 5x - 6x^2 - 7x^3}{2x^2}})dx
…
The text I am using has no answers and when I tried to use the integral calculator at http://www.numberempire.com/integralcalculator.php I get an answer that is a fraction.

(use "\left(" and "\right)" for big brackets )

??

I used that calculator and got

(10*x*log(x)-7*x^3-12*x^2-6)/(4*x)

 

[username12345]

I used that calculator and got

(10*x*log(x)-7*x^3-12*x^2-6)/(4*x)

I got that too but I don't know where I went wrong with my calculation...

 

[arildno]

Yes, your answer is correct.

It is also equivalent to that fraction, as far as I can see.

 

[username12345]

Great.

And thanks for the tip on the brackets.