Question

If $\int \frac{x^2-x-1}{\sqrt[3]{31-3x}}dx=\frac{-(x^2-x-1){(1-3x)}^{\frac{2}{3}}}{2}-\frac{(2x-1){(1-3x)}^{\frac{8}{3}}}{k}+C$, where $k=$

Answer 1

$I=\int \frac{x^2-x-1}{\sqrt[3]{31-3x}}dx=\int (x^2-x-1){(1-3x)}^{\frac{-1}{3}}dx$

$=(x^2-x-1)\frac{{(1-3x)}^{\frac{2}{3}}}{\left(\frac{2}{3}\right)(-3)}-\int \frac{{(1-3x)}^{\frac{2}{3}}}{\left(\frac{2}{3}\right)(-3)}.(2x-1)dx$

$=\frac{(x^2-x-1){(1-3x)}^{\frac{2}{3}}}{-2}+\frac{1}{2}\int (2x-1){(1-3x)}^{\frac{2}{3}}dx$

$=\frac{(x^2-x-1){(1-3x)}^{\frac{2}{3}}}{-2}+\frac{1}{2}(2x-1)\frac{{(1-3x)}^{\frac{5}{3}}}{\left(\frac{5}{3}\right)(-3)}-\frac{1}{2}\int \frac{{(1-3x)}^{\frac{5}{3}}}{\left(\frac{5}{3}\right)(-3)}$

$=\frac{(x^2-x-1){(1-3x)}^{\frac{2}{3}}}{-2}+\frac{(2x-1){(1-3x)}^{\frac{5}{3}}}{-10}+\frac{1}{5}\frac{{(1-3x)}^{\frac{8}{3}}}{\left(\frac{8}{3}\right)(-3)}+C$

$=\frac{(x^2-x-1){(1-3x)}^{\frac{2}{3}}}{-2}+\frac{(2x-1){(1-3x)}^{\frac{5}{3}}}{-10}+\frac{{(1-3x)}^{\frac{8}{3}}}{-40}+C$
$\therefore k=40$