Question

find the indefinite integral.
int sqrt{1 + 4 x ^ { 2 } } ( 8 x ) d x
int sqrt { 1 + 4 x ^ { 2 } } ( 8 x ) d x =

Explanation:

Step1: Use substitution

Let \(u = 1 + 4x^{2}\), then \(du=8xdx\).

Step2: Integrate with respect to \(u\)

The integral \(\int\sqrt{1 + 4x^{2}}(8x)dx=\int\sqrt{u}du\).
Since \(\int u^{n}du=\frac{u^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), for \(n=\frac{1}{2}\), we have \(\int u^{\frac{1}{2}}du=\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C=\frac{2}{3}u^{\frac{3}{2}}+C\).

Step3: Substitute back \(u = 1 + 4x^{2}\)

Substituting \(u = 1+4x^{2}\) back into the result, we get \(\frac{2}{3}(1 + 4x^{2})^{\frac{3}{2}}+C\).

Answer:

\(\frac{2}{3}(1 + 4x^{2})^{\frac{3}{2}}+C\)