Can someone help me integrate:
∫ x^3/X^2+1 dx from 1 to 2
Thank you
Observe that $\displaystyle (x^2 + 1)(x + 1) = x^3 + x^2 + x + 1$.
Therefore,
$\displaystyle \int \frac{x^3}{x^2 + 1}dx =$
$\displaystyle \int \frac{x^3 + x^2 + x + 1 - (x^2 + 1) - x}{x^2 + 1}dx =$
$\displaystyle \int \frac {(x^2 + 1)(x + 1)}{x^2 + 1} - \frac{x^2 + 1}{x^2 + 1} - \frac {x}{x^2 + 1}dx =$
$\displaystyle \int x + 1 - 1 - \frac{x}{x^2 + 1}dx =$
$\displaystyle \int x - \frac{x}{x^2 + 1}dx =$
$\displaystyle \int x dx - \int \frac{x}{x^2 + 1} dx$
Here's an alternative method (you'll get the same result).
Let $\displaystyle u=x^2+1$ so that $\displaystyle du=2x \, dx$.
Then we have: $\displaystyle u-1 = x^2$ and $\displaystyle \frac{du}{2} = x \, dx$.
So:
$\displaystyle \int_1^2 \frac{x^3}{x^2+1} \, dx ~=~ \int_1^2 \frac{x^2}{x^2+1} \, x \, dx ~=~ \int_2^5 \frac{u-1}{u} \, \frac{du}{2} ~=~ \frac{1}{2} \int_2^5 \left( 1 - \frac{1}{u} \right) du$