Thread: Integration Problem

$\displaystyle \int_{0}^{\sqrt3} \frac{5x^2 dx}{x^2 +1}$

Not entirely sure how to do this one.

Originally Posted by looseenz2

$\displaystyle \int_{0}^{\sqrt3} \frac{5x^2 dx}{x^2 +1}$

Not entirely sure how to do this one.

Note that $\displaystyle \frac{5x^2}{x^2 + 1} = 5 \left( \frac{x^2}{x^2 + 1}\right) = 5 \left( \frac{(x^2 + 1) - 1}{x^2 + 1}\right) = 5 - \frac{5}{x^2 + 1}$.

Originally Posted by mr fantastic

Note that $\displaystyle \frac{5x^2}{x^2 + 1} = 5 \left( \frac{x^2}{x^2 + 1}\right) = 5 \left( \frac{(x^2 + 1) - 1}{x^2 + 1}\right) = 5 - \frac{5}{x^2 + 1}$.

Could you please explain what you did between the last two steps? I'm not entirely sure what you did.

Originally Posted by looseenz2

Could you please explain what you did between the last two steps? I'm not entirely sure what you did.

the basic rule is ...

when the numerator has a degree $\displaystyle \ge$ the degree of the denominator, divide the fraction.

this can be done with long division (as shown below), or the "shortcut" shown by mr. F

Code:

........1
      ----------
x^2+1 | x^2 
........x^2 + 1
.....  -----------
........... - 1

the quotient is $\displaystyle 1 + \frac{-1}{x^2+1} = 1 - \frac{1}{x^2+1}
$