Roboguru

Hasil dari  adalah....

Pertanyaan

Hasil dari begin mathsize 14px style integral left parenthesis 4 x minus 6 right parenthesis square root of left parenthesis x squared minus 3 x plus 5 right parenthesis end root space d x end style adalah....undefined 

  1. begin mathsize 14px style 8 over 3 left parenthesis x squared minus 3 x plus 5 right parenthesis square root of left parenthesis x squared minus 3 x plus 5 right parenthesis end root plus C end style  

  2. begin mathsize 14px style 4 over 3 left parenthesis x squared minus 3 x plus 5 right parenthesis square root of left parenthesis x squared minus 3 x plus 5 right parenthesis end root plus C end style  

  3. begin mathsize 14px style 2 over 3 left parenthesis x squared minus 3 x plus 5 right parenthesis square root of left parenthesis x squared minus 3 x plus 5 right parenthesis end root plus C end style  

  4. begin mathsize 14px style 1 half left parenthesis x squared minus 3 x plus 5 right parenthesis square root of left parenthesis x squared minus 3 x plus 5 right parenthesis end root plus C end style  

  5. begin mathsize 14px style 1 third left parenthesis x squared minus 3 x plus 5 right parenthesis square root of left parenthesis x squared minus 3 x plus 5 right parenthesis end root plus C end style 

Pembahasan Soal:

Misalkan begin mathsize 14px style u equals x squared minus 3 x plus 5 space end style, maka begin mathsize 14px style fraction numerator d u over denominator d x end fraction equals 2 x minus 3 space atau space fraction numerator d straight u over denominator 2 straight x minus 3 end fraction equals straight d x end style sehingga 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral open parentheses 4 x minus 6 close parentheses square root of open parentheses x squared minus 3 x plus 5 close parentheses space end root d x end cell equals cell integral 2 open parentheses 2 x minus 3 close parentheses open parentheses x squared minus 3 x plus 5 close parentheses to the power of 1 half end exponent d x end cell row blank equals cell integral 2 open parentheses 2 x minus 3 close parentheses u to the power of 1 half end exponent fraction numerator d u over denominator 2 x minus 3 end fraction end cell row blank equals cell integral 2 u to the power of 1 half end exponent d u end cell row blank equals cell 2 times 2 over 3 u to the power of 3 over 2 end exponent plus C end cell row blank equals cell 4 over 3 u square root of u plus C end cell row blank equals cell 4 over 3 open parentheses x squared minus 3 x plus 5 close parentheses square root of x squared minus 3 x plus 5 end root plus C end cell end table end style

Jadi, jawaban yang tepat adalah B.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 20 Mei 2021

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