Question

integral of 4(6x-1)^2/3 dx

Answer 1

$\displaystyle\int4(6x-1)^{2/3}\,\mathrm dx$

Let $y=6x-1$, so that $\mathrm dy=6\,\mathrm dx\implies\dfrac{\mathrm dy}6=\mathrm dx$. Then

$\displaystyle\int4(6x-1)^{2/3}\,\mathrm dx=\int\frac46y^{2/3}\,\mathrm dy$

Simplifying and applying the power rule gives

$\displaystyle\frac23\frac{y^{5/3}}{\frac53}+C=\frac23\times\frac35y^{5/3}+C=\frac25y^{5/3}+C$

and back-substituting to get this in terms of $x$, you end up with

$\dfrac25(6x-1)^{5/3}+C$