Question

find the antiderivative of the given derivative.
\\( \frac { d p } { d x } = e ^ { x } + 6 x - 6 x
\\( p = \square
\\( \frac { e ^ { x } - e ^ { - 6 x } } { 3 } \\)

Explanation:

Step1: Integrate \(e^{x}\)

$$\int e^{x}dx = e^{x}$$

Step2: Integrate \(6x\)

Using the power - rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), for \(n = 1\) in \(6x\) (i.e., \(\int6x dx=6\int xdx\)), we have \(6\times\frac{x^{2}}{2}=3x^{2}\)

Step3: Integrate \(-\frac{e^{6x}}{3}\)

Let \(u = 6x\), then \(du=6dx\) and \(dx=\frac{1}{6}du\). So \(\int-\frac{e^{6x}}{3}dx=-\frac{1}{3}\times\frac{1}{6}\int e^{u}du=-\frac{1}{18}e^{6x}\)

Answer:

\(p = e^{x}+3x^{2}-\frac{1}{18}e^{6x}+C\)