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f ″ ( x ) = 3 ( x + 4 ) ⋅ 2 ( x + 4 ) 6 = 6 ( x + 4 ) 5 {\displaystyle f''(x)={\frac {3(x+4)\cdot 2}{(x+4)^{6}}}={\frac {6}{(x+4)^{5}}}}
f ′ ( x ) = − 2 ( x + 4 ) ( x + 4 ) 4 = − 2 ( x + 4 ) 3 {\displaystyle f'(x)=-{\frac {2(x+4)}{(x+4)^{4}}}=-{\frac {2}{(x+4)^{3}}}}
f ( x ) = 1 x 2 + 8 x + 16 = 1 ( x + 4 ) 2 {\displaystyle f(x)={\frac {1}{x^{2}+8x+16}}={\frac {1}{(x+4)^{2}}}}
∫ f ( x ) d x = − 1 x + 4 {\displaystyle \int f(x)dx=-{\frac {1}{x+4}}}
∫ ∫ f ( x ) d x ⋅ d x = − log ( x + 4 ) {\displaystyle \int \int f(x)dx\cdot dx=-\log(x+4)}
