Question

find the indefinite integral.

$$ int \frac { 1 } { 2 + 4 x ^ { 4 } } ( 16 x ^ { 3 } ) d x $$

$$ int \frac { 1 } { 2 + 4 x ^ { 4 } } ( 16 x ^ { 3 } ) d x = $$

Explanation:

Step1: Let \(u = 2 + 4x^{4}\)

Differentiate \(u\) with respect to \(x\): \(du=(0 + 16x^{3})dx=16x^{3}dx\)

Step2: Substitute \(u\) and \(du\) into the integral

The integral \(\int\frac{1}{2 + 4x^{4}}(16x^{3})dx\) becomes \(\int\frac{1}{u}du\)

Step3: Integrate \(\int\frac{1}{u}du\)

Using the formula \(\int\frac{1}{u}du=\ln|u|+C\) (where \(C\) is the constant of integration), we get \(\ln|u|+C\)

Step4: Substitute back \(u = 2 + 4x^{4}\)

The result is \(\ln(2 + 4x^{4})+C\) (since \(2 + 4x^{4}>0\) for all real \(x\), the absolute - value can be removed)

Answer:

\(\ln(4x^{4}+2)+C\)