$\int \frac{12 u^{11}}{u^4\+u^3} \mathit{\ⅆ}u$

$\=12\int \frac{u^8}{u\+1} \mathit{\ⅆ}u$

$\=12\int \left(u^7-u^6\+u^5-u^4\+u^3-u^2\+u-1\+\frac{1}{u\+1}\right) \mathit{\ⅆ}u$

$\=12\(\frac{u^8}{8}-\frac{u^7}{7}\+\frac{u^6}{6}-\frac{u^5}{5}\+\frac{u^4}{4}-\frac{u^3}{3}\+\frac{u^2}{2}-u\+\ln \(\left|u\+1\right|\)$

$\=12\(\frac{x^{3\/4}}{8}-\frac{x^{7\/12}}{7}\+\frac{\sqrt{x}}{6}-\frac{x^{5\/12}}{5}\+\frac{x^{1\/3}}{4}-\frac{x^{1\/4}}{3}\+\frac{x^{1\/6}}{2}-x^{1\/12}\+\ln \(\left|x^{1\/12}\+1\right|\)$

$\begin{array}{c}\int \frac{1}{x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}\ⅆx\hfill \\ & \text{=}& \int \left(12u^7-12u^6+12u^5-12u^4+12u^3-12u^2+12u-12+\frac{12}{u+1}\right)\ⅆu\hfill & \left[\mathrm{change}\,x=u^{12}\,u\right]\end{array}$

Table 6.6.2(a)   First step taken in the Integration Methods tutor

$\mathrm{with}\left(\mathrm{IntegrationTools}\right)\:$

$Q≔\int \frac{1}{x^{1\/3}\+x^{1\/4}} \mathit{\ⅆ}x\:$

$q_1≔\mathrm{simplify}\left(\mathrm{Change}\left(Q\,x\=u^{12}\,u\right)\right) assuming u\>0$

$12\left(\∫\frac{u^8}{u\+1}\ⅆu\right)$

$q_2≔12\cdot \mathrm{Int}\left(\mathrm{convert}\left(\mathrm{GetIntegrand}\left(q_1\right)\,\mathrm{parfrac}\,u\right)\,u\right)$

$12\left(\∫\left(u^7-u^6\+u^5-u^4\+u^3-u^2\+u-1\+\frac{1}{u\+1}\right)\ⅆu\right)$

$q_3≔\mathrm{value}\left(q_2\right)$

$\frac{3}{2}{u}^8-\frac{12}{7}{u}^7\+2u^6-\frac{12}{5}{u}^5\+3u^4-4u^3\+6u^2-12u\+12\ln \left(u\+1\right)$

$\mathrm{eval}\left(q_3\,u\=x^{1\/12}\right)$

$\frac{3}{2}{x}^{2\/3}-\frac{12}{7}{x}^{7\/12}\+2\sqrt{x}-\frac{12}{5}{x}^{5\/12}\+3x^{1\/3}-4x^{1\/4}\+6x^{1\/6}-12x^{1\/12}\+12\ln \left(x^{1\/12}\+1\right)$

Table 6.6.2(b)   Stepwise solution based on commands from the IntegrationTools package