Question

find the indefinite integral.
int sqrt{1 + 4 x ^ { 4 } } left( 16 x ^ { 3 }
ight) d x
int sqrt { 1 + 4 x ^ { 4 } } left( 16 x ^ { 3 }
ight) d x =

Explanation:

Step1: Set substitution

Let \(u = 1 + 4x^{4}\). Then, find the derivative \(du\).
Using the power rule \(\frac{d}{dx}(ax^{n})=nax^{n - 1}\), we have \(\frac{du}{dx}=16x^{3}\), so \(du = 16x^{3}dx\).

Step2: Substitute into the integral

The integral \(\int\sqrt{1 + 4x^{4}}(16x^{3})dx\) becomes \(\int\sqrt{u}du\).
Since \(\sqrt{u}=u^{\frac{1}{2}}\), the integral is \(\int u^{\frac{1}{2}}du\).

Step3: Integrate

Using the power - rule for integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), for \(n=\frac{1}{2}\), we get \(\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C=\frac{u^{\frac{3}{2}}}{\frac{3}{2}}+C=\frac{2}{3}u^{\frac{3}{2}}+C\).

Step4: Back - substitute

Substitute \(u = 1 + 4x^{4}\) back into the expression. So the integral is \(\frac{2}{3}(1 + 4x^{4})^{\frac{3}{2}}+C\).

Answer:

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