$\int\frac{\left(4x+6\right)}{\left(x^2+3x-8\right)}dx$
$4x^2-24x-189$
$\int tan^6\:2x\:dx$
$6x^2+51x+81=9x^2+81x+153$
$x^2+7x=98$
$\int-\frac{2}{\left(x-1\right)\left(x+1\right)+1}dx$
$\frac{t^{2}-6t-7}{t^{2}-81}$
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