- #1
shamieh
- 539
- 0
Suppose:
\(\displaystyle \int^{\infty}_1 \frac{x}{x^2 + 1} \, dx\)
I can't do this can I?
Let \(\displaystyle u = x^2 + 1\)
so \(\displaystyle du = 2x \, dx\)
\(\displaystyle \therefore \frac{du}{2} = x \, dx\) ?
\(\displaystyle \int^{\infty}_1 \frac{x}{x^2 + 1} \, dx\)
I can't do this can I?
Let \(\displaystyle u = x^2 + 1\)
so \(\displaystyle du = 2x \, dx\)
\(\displaystyle \therefore \frac{du}{2} = x \, dx\) ?