Question

Find the integral [(3x-2)^4]/[7]

Answer 1

Is there an easier way to expand it?

Answer 2

\[\frac{ (3x-2)^{4} }{ 7 } dx\]

Answer 3

yeah try \(u=3x-2,du=3dx,\frac{1}{3}u=dx\)

Answer 4

you get
\[\frac{1}{21}\int u^4du\] and

Answer 5

Ah! yes okay thanks!!!!!!

Answer 6

so would it be
\[\frac{ (3x-2)^{5} }{ 135 } +c\]

Answer 7

i think your denominator is wrong

Answer 8

\(21\times 5=105\)

Answer 9

oh haha I wrote down 27 for some reason

Answer 10

∫(3x-2)^4/[7] dx
1/7∫(3x-2)^5dx

so,
1/7{(3x-2)^5/(5*3)}+c therefore dinominator '3' is derivative of base.

finally;

1/135(3x-2)^5 +c .