# Integral of fraction - is this correct?

It looks much better now.In summary, the conversation is about finding the integral of a given expression. The process used to solve it involved splitting the fraction, simplifying it, and then adding the resulting integrals. The final answer is -\frac{3}{2x} + \frac{5}{2}\ln x - 3x - \frac{7}{4}x^2 + c. The use of an online integral calculator also resulted in a correct answer, which is equivalent to the original one.

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.

Last edited:
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)

tiny-tim said:
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...