# Thread: Need help with definite integral evaluation

1. ## Need help with definite integral evaluation

Can someone help me integrate:

∫ x^3/X^2+1 dx from 1 to 2

Thank you

2. Once you integrate you get x^2/2+x
substitude x'es with 2 and the result minus the substitution of 1
which equals to 5/2 asuming that the "+1" is not in the denominator

Can someone help me integrate:

∫ x^3/X^2+1 dx from 1 to 2

Thank you

3. Observe that $(x^2 + 1)(x + 1) = x^3 + x^2 + x + 1$.

Therefore,

$\int \frac{x^3}{x^2 + 1}dx =$

$\int \frac{x^3 + x^2 + x + 1 - (x^2 + 1) - x}{x^2 + 1}dx =$

$\int \frac {(x^2 + 1)(x + 1)}{x^2 + 1} - \frac{x^2 + 1}{x^2 + 1} - \frac {x}{x^2 + 1}dx =$

$\int x + 1 - 1 - \frac{x}{x^2 + 1}dx =$

$\int x - \frac{x}{x^2 + 1}dx =$

$\int x dx - \int \frac{x}{x^2 + 1} dx$

4. Here's an alternative method (you'll get the same result).

Let $u=x^2+1$ so that $du=2x \, dx$.

Then we have: $u-1 = x^2$ and $\frac{du}{2} = x \, dx$.

So:

$\int_1^2 \frac{x^3}{x^2+1} \, dx ~=~ \int_1^2 \frac{x^2}{x^2+1} \, x \, dx ~=~ \int_2^5 \frac{u-1}{u} \, \frac{du}{2} ~=~ \frac{1}{2} \int_2^5 \left( 1 - \frac{1}{u} \right) du$