5.4 Indefinite Integrals and the Net Change Theorem/39

$\int _{1}^{64}{\frac {1+{\sqrt[{3}]{x}}}{\sqrt {x}}}dx$

= $\int _{1}^{64}{\frac {1}{x^{1/2}}}$ + $\int _{1}^{64}{\frac {x^{1/3}}{x^{1/2}}}$

= $\int _{1}^{64}x^{-1/2}+x^{1/3-1/2}$ = $\int _{1}^{64}x^{-1/2}+x^{-1/6}$

Add one to the exponents and divide by the new exponent

= $\int _{1}^{64}{\frac {x^{1/2}}{\frac {1}{2}}}+{\frac {x^{5/6}}{\frac {5}{6}}}$ = $\int _{1}^{64}2x^{\frac {1}{2}}+{\frac {6}{5}}x^{\frac {5}{6}}$

=$2(x)^{\frac {1}{2}}+{\frac {6}{5}}(x)^{\frac {5}{6}}{\bigg |}_{1}^{64}$

= $2(64)^{\frac {1}{2}}+{\frac {6}{5}}(64)^{\frac {5}{6}}-(2(1)^{\frac {1}{2}}+{\frac {6}{5}}(1)^{\frac {5}{6}})$

= $16+38.4-(2+1.2)$

= $54.4-3.2$

= $51.2$

= ${\frac {256}{5}}$

$\int {}{}1+tan^{2}x*dx={\frac {d}{dx}}(tan(x))$

${\frac {d}{dx}}(tan(x))={\frac {d}{dx}}\left({\frac {sin(x)}{cos(x)}}\right)={\frac {cos^{2}x-(-(sin(x))(sin(x))}{cos^{2}(x)}}$

$={\frac {cos^{2}x+sin^{2}(x))}{cos^{2}(x)}}={\frac {cos^{2}x}{cos^{2}(x)}}+{\frac {sin^{2}x}{cos^{2}(x)}}$

$=1+tan^{2}(x)$