Question

∫xe^{4x^{2}+1}dx

Explanation:

Step1: Substitution

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

Step2: Integral transformation

The original integral \(\int xe^{4x^{2}+1}dx\) becomes \(\frac{1}{8}\int e^{u}du\).

Step3: Integrate \(e^{u}\)

Since \(\int e^{u}du = e^{u}+C\), then \(\frac{1}{8}\int e^{u}du=\frac{1}{8}e^{u}+C\).

Step4: Back - substitution

Substitute \(u = 4x^{2}+1\) back, we get \(\frac{1}{8}e^{4x^{2}+1}+C\).

Answer:

\(\frac{1}{8}e^{4x^{2}+1}+C\)