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∫ 3 6 x − 5 ​ 4 ​ d x = ...

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  1. begin mathsize 14px style 5 cube root of open parentheses 6 x minus 5 close parentheses squared end root plus C end style 

  2. begin mathsize 14px style 4 cube root of open parentheses 6 x minus 5 close parentheses squared end root plus C end style 

  3. begin mathsize 14px style 3 cube root of open parentheses 6 x minus 5 close parentheses squared end root plus C end style 

  4. begin mathsize 14px style 2 cube root of open parentheses 6 x minus 5 close parentheses squared end root plus C end style 

  5. begin mathsize 14px style cube root of open parentheses 6 x minus 5 close parentheses squared end root plus C end style 

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jawaban yang benar adalah E.

jawaban yang benar adalah E.

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Gunakan konsep integral substitusi misalkan maka: kemudian substitusikan: Dengan demikian, . Oleh karena itu, jawaban yang benar adalah E.

Gunakan konsep integral substitusi misalkan begin mathsize 14px style straight u equals 6 x minus 5 end style maka:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row straight u equals cell 6 x minus 5 end cell row cell fraction numerator du over denominator straight d x end fraction end cell equals 6 row cell straight d x end cell equals cell 1 over 6 du end cell end table end style 

kemudian substitusikan:

table attributes columnalign right center left columnspacing 0px end attributes row cell integral fraction numerator 4 over denominator cube root of 6 x minus 5 end root end fraction straight d x end cell equals cell integral 4 over open parentheses 6 x minus 5 close parentheses to the power of begin display style 1 third end style end exponent straight d x end cell row blank equals cell integral 4 open parentheses 6 x minus 5 close parentheses to the power of negative 1 third end exponent straight d x end cell row blank equals cell integral 4 straight u to the power of negative 1 third end exponent 1 over 6 du end cell row blank equals cell 4 over 6 integral straight u to the power of negative 1 third end exponent space du end cell row blank equals cell 4 over 6 open square brackets fraction numerator 1 over denominator negative begin display style 1 third end style plus 1 end fraction times straight u to the power of negative 1 third plus 1 end exponent close square brackets plus C end cell row blank equals cell 4 over 6 open square brackets fraction numerator 1 over denominator begin display style 2 over 3 end style end fraction times straight u to the power of 2 over 3 end exponent close square brackets plus C end cell row blank equals cell 4 over 6 open square brackets 3 over 2 times straight u to the power of 2 over 3 end exponent close square brackets plus C end cell row blank equals cell 4 over 6 times 3 over 2 times straight u to the power of 2 over 3 end exponent plus C end cell row blank equals cell straight u to the power of 2 over 3 end exponent plus C end cell row blank equals cell cube root of open parentheses 6 x minus 5 close parentheses squared end root plus C end cell end table

Dengan demikian, begin mathsize 14px style integral fraction numerator 4 over denominator cube root of 6 x minus 5 end root end fraction space straight d x equals cube root of open parentheses 6 x minus 5 close parentheses squared end root plus C end style.

Oleh karena itu, jawaban yang benar adalah E.

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