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Integral of fraction - is this correct?

  • #1
I am trying to find the integral of the following:

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

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

[tex]\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 [/tex]

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

[tex]-\frac{3}{2x} + \frac{5}{2}\ln x - 3x - \frac{7}{4}x^2 + c[/tex]

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:

Answers and Replies

  • #2
tiny-tim
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Homework Helper
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I am trying to find the integral of the following:

[tex]\int({\frac{3 + 5x - 6x^2 - 7x^3}{2x^2}})dx[/tex]

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 :wink:)

?? :confused:

I used that calculator and got

(10*x*log(x)-7*x^3-12*x^2-6)/(4*x)
 
  • #3
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...
 
  • #4
arildno
Science Advisor
Homework Helper
Gold Member
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Yes, your answer is correct.

It is also equivalent to that fraction, as far as I can see.
 
  • #5
Great.

And thanks for the tip on the brackets.
 

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