Using the substitution \(u = \sqrt{x}\), find the exact value of \(\int_{3}^{\infty} \frac{1}{(x+1)\sqrt{x}} \, dx\).
Solution
Let \(u = \sqrt{x}\), then \(x = u^2\) and \(dx = 2u \, du\).
The limits change from \(x = 3\) to \(u = \sqrt{3}\) and \(x = \infty\) to \(u = \infty\).
Substitute into the integral:
\(\int_{\sqrt{3}}^{\infty} \frac{1}{(u^2 + 1)u} \, 2u \, du = \int_{\sqrt{3}}^{\infty} \frac{2}{u^2 + 1} \, du\).
The integral \(\int \frac{2}{u^2 + 1} \, du\) is \(2 \arctan(u)\).
Evaluate from \(\sqrt{3}\) to \(\infty\):
\(2 \left[ \arctan(\infty) - \arctan(\sqrt{3}) \right] = 2 \left[ \frac{\pi}{2} - \frac{\pi}{3} \right]\).
\(= 2 \left[ \frac{\pi}{6} \right] = \frac{1}{3} \pi\).
Log in to record attempts.
