4. $int_{0}^{infty}\frac{1}{sqrt4{1 + x}}dx$

Explanation:

Step1: Substitution

Let (u = 1 + x), then (du=dx). When (x = 0), (u = 1); as (x\to\infty), (u\to\infty). The integral becomes (\int_{1}^{\infty}\frac{1}{\sqrt[4]{u}}du=\int_{1}^{\infty}u^{-\frac{1}{4}}du).

Step2: Integrate

Using the power - rule (\int u^{n}du=\frac{u^{n + 1}}{n+1}+C(n\neq - 1)), for (n=-\frac{1}{4}), we have (\lim_{b\to\infty}\left[\frac{u^{-\frac{1}{4}+1}}{-\frac{1}{4}+1}\right]{1}^{b}=\lim{b\to\infty}\left[\frac{u^{\frac{3}{4}}}{\frac{3}{4}}\right]_{1}^{b}).

Step3: Evaluate the limit

(\lim_{b\to\infty}\frac{4}{3}\left(b^{\frac{3}{4}}-1^{\frac{3}{4}}\right)). Since (\lim_{b\to\infty}b^{\frac{3}{4}}=\infty), the value of the integral is (\infty).

Answer:

The integral (\int_{0}^{\infty}\frac{1}{\sqrt[4]{1 + x}}dx) diverges.