Using Properties of the Indefinite Integral

1. \[ \begin{eqnarray*} \int \left( \dfrac{12}{x^{5}}+\dfrac{1}{\sqrt{x}}\right) dx&=&12\int \dfrac{1}{x^{5}}\,dx+\int \dfrac{1}{\sqrt{x}}\,dx=12\int x^{-5}\,dx+\int x^{-1/2}\,dx \\ &=& 12\left( \dfrac{x^{-4}}{{-}4}\right) +\dfrac{ x^{1/2}}{\dfrac{1}{2}}+C = -\dfrac{3}{x^{4}}+2\sqrt{x}+C \end{eqnarray*} \]
2. \[ \begin{eqnarray*} \int \dfrac{x^{2}+6}{x^{2}+1} dx & \underset{\underset{{\color{#0066A7}{\hbox{Algebra}}}}{\color{#0066A7}{\uparrow}}}{=}& \int \dfrac{( x^{2}+1) +5}{x^{2}+1} dx =\int \left[ \dfrac{x^{2}+1}{x^{2}+1}+\dfrac{5}{x^{2}+1}\right] dx \\ &=&\int \left[ 1+\dfrac{5}{x^{2}+1}\right] dx \underset{\underset{{\color{#0066A7}{\hbox{Sum Property}}}}{\color{#0066A7}{\uparrow }}} {=} \int dx+\int \dfrac{5}{x^{2}+1} dx \\ &=&\int dx+5\int \dfrac{1}{x^{2}+1} dx=x+5 \tan ^{-1}x+C \end{eqnarray*} \]