Question

question 4
find the indefinite integral.
$\int 3 x ( 4 x ^ { 2 } + 3 ) ^ { 6 } d x$
$\bigcirc x ^ { 3 } ( 4 x ^ { 2 } + 3 ) ^ { 7 } + c$
$\bigcirc \frac { 3 } { 56 } ( 4 x ^ { 2 } + 3 ) ^ { 7 } + c$
$\bigcirc \frac { 3 } { 2 } ( 4 x ^ { 2 } + 3 ) ^ { 7 } + c$
$\bigcirc 3 x ^ { 2 } ( 4 x ^ { 2 } + 3 ) ^ { 7 } + c$

Explanation:

Step1: Use substitution

Let \(u = 4x^{2}+3\), then \(du=8xdx\), and \(xdx=\frac{1}{8}du\). The integral \(\int3x(4x^{2}+3)^{6}dx\) becomes \(\frac{3}{8}\int u^{6}du\).

Step2: Integrate \(u^{6}\)

Using the power - rule \(\int u^{n}du=\frac{u^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), for \(n = 6\), we have \(\frac{3}{8}\times\frac{u^{7}}{7}+C=\frac{3u^{7}}{56}+C\).

Step3: Substitute back \(u = 4x^{2}+3\)

We get \(\frac{3(4x^{2}+3)^{7}}{56}+C\).

Answer:

\(\frac{3}{56}(4x^{2}+3)^{7}+C\) (the second option)