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Hasil ∫3x2(2x3+5)​dx=...

Pertanyaan

Hasil integral 3 x squared square root of left parenthesis 2 x cubed plus 5 right parenthesis space end root d x equals...

  1. 3 over 4 open parentheses 2 x cubed plus 5 close parentheses square root of left parenthesis 2 x cubed plus 5 end root right parenthesis plus c

  2. 1 half open parentheses 2 x cubed plus 5 close parentheses square root of left parenthesis 2 x cubed plus 5 right parenthesis end root plus c

  3. 2 over 5 open parentheses 2 x cubed plus 5 close parentheses square root of left parenthesis 2 x cubed plus 5 right parenthesis end root plus c

  4. 1 third open parentheses 2 x cubed plus 5 close parentheses square root of left parenthesis 2 x cubed plus 5 right parenthesis end root plus c

  5. 1 over 6 open parentheses 2 x cubed plus 5 close parentheses square root of left parenthesis 2 x cubed plus 5 right parenthesis end root plus c

Pembahasan Soal:

Misal :

U space equals 2 x cubed space plus 5  d u space equals space 6 x squared d x  d x equals fraction numerator 1 over denominator 6 x squared end fraction d u equals fraction numerator 1 over denominator 2 left parenthesis 3 x squared right parenthesis end fraction d u  integral 3 x squared left parenthesis 2 x cubed plus 5 right parenthesis to the power of 1 half end exponent d x equals integral 3 x squared. U to the power of 1 half end exponent fraction numerator 1 over denominator 2 left parenthesis 3 x squared right parenthesis end fraction d u  equals 1 half.2 over 3 U to the power of 3 over 2 end exponent plus c  equals 1 third left parenthesis 2 x squared plus 5 right parenthesis to the power of 3 over 2 end exponent plus c  equals 1 third open parentheses 2 x cubed plus 5 close parentheses square root of 2 x cubed plus 5 end root plus c

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Mustikowati

Mahasiswa/Alumni Universitas Negeri Jakarta

Terakhir diupdate 04 Oktober 2021

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Pertanyaan yang serupa

∫x(2x2−5)3dx adalah ....

Pembahasan Soal:

Permasalahan integral pada soal akan diselesaikan dengan metode integral subtitusi.

Misalkan u equals 2 x squared minus 5, maka diperoleh fraction numerator d u over denominator d x end fraction equals 4 x sehingga fraction numerator d u over denominator 4 end fraction equals x d x.

Dengan menyubtitusi variable u dan fraction numerator d u over denominator 4 end fraction ke integral pada soal, diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell integral x open parentheses 2 x squared minus 5 close parentheses cubed d x end cell equals cell integral open parentheses 2 x squared minus 5 close parentheses cubed x d x end cell row blank equals cell integral open parentheses u close parentheses cubed fraction numerator d u over denominator 4 end fraction end cell end table

Oleh karena integral x to the power of n d x equals fraction numerator 1 over denominator n plus 1 end fraction x to the power of n plus 1 end exponent plus C, maka penyelesaian integral di atas adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell integral open parentheses u close parentheses cubed fraction numerator d u over denominator 4 end fraction end cell equals cell 1 fourth integral u cubed d u end cell row blank equals cell 1 fourth open parentheses fraction numerator 1 over denominator 3 plus 1 end fraction u to the power of 3 plus 1 end exponent plus C close parentheses end cell row blank equals cell 1 fourth open parentheses 1 fourth u to the power of 4 plus C close parentheses end cell row blank equals cell 1 over 16 u to the power of 4 plus 1 fourth C end cell row blank equals cell 1 over 16 u to the power of 4 plus C end cell end table

Dengan menyubtitusi kembali variable u ke hasil pengintegralan di atas, diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 over 16 u to the power of 4 plus C end cell equals cell 1 over 16 open parentheses 2 x squared minus 5 close parentheses to the power of 4 plus C end cell end table

Jadi, integral x open parentheses 2 x squared minus 5 close parentheses cubed d x equals 1 over 16 open parentheses 2 x squared minus 5 close parentheses to the power of 4 plus C.space 

0

Roboguru

Hasil dari  adalah ….

Pembahasan Soal:

Akan dicari hasil dari

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Perhatikan bahwa

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell d over dx open square brackets open parentheses 4 x plus 1 close parentheses to the power of 6 close square brackets end cell equals cell 6 ⋅ open parentheses 4 x plus 1 close parentheses to the power of 5 ⋅ open parentheses 4 x plus 1 close parentheses to the power of apostrophe end cell row blank equals cell 6 ⋅ open parentheses 4 x plus 1 close parentheses to the power of 5 ⋅ 4 end cell row blank equals cell 24 open parentheses 4 x plus 1 close parentheses to the power of 5 end cell end table end style    

Akibatnya, kita peroleh   

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral 24 left parenthesis 4 x plus 1 right parenthesis to the power of 5 space dx end cell equals cell integral fraction numerator straight d over denominator dx end fraction left square bracket left parenthesis 4 straight x plus 1 right parenthesis to the power of 6 right square bracket space dx end cell row cell integral 24 left parenthesis 4 straight x plus 1 right parenthesis to the power of 5 space dx end cell equals cell left parenthesis 4 straight x plus 1 right parenthesis to the power of 6 plus straight C end cell end table end style

Sehingga, perhatikan bahwa

 Error converting from MathML to accessible text. 

Jadi, jawaban yang tepat adalah B.

0

Roboguru

Hasil dari ∫(3x+8)3dx adalah ....

Pembahasan Soal:

Dimisalkan u equals 3 x plus 8 maka

d u equals 3 space d x d x equals 1 third space d u 

Bentuk integral di atas dapat diselesaikan dengan substitusi :

table attributes columnalign right center left columnspacing 0px end attributes row cell integral open parentheses 3 x plus 8 close parentheses cubed d x end cell equals cell 1 third integral u cubed space d u end cell row blank equals cell 1 third open square brackets 1 fourth times u to the power of 4 plus C close square brackets end cell row blank equals cell 1 over 12 left parenthesis 3 x plus 8 right parenthesis to the power of 4 plus C end cell end table 

Dengan demikian, hasil dari integral open parentheses 3 x plus 8 close parentheses cubed d x adalah 1 over 12 left parenthesis 3 x plus 8 right parenthesis to the power of 4 plus c.

Jadi, jawaban yang tepat adalah A.

0

Roboguru

∫36x−5​4​dx=...

Pembahasan Soal:

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.

0

Roboguru

Hasildari∫(2x−1)x2−x+5​dx=....

Pembahasan Soal:

Misalkan space straight u equals straight x squared minus straight x plus 5  Akibatnya space du over dx equals 2 straight x minus 1 space atau space dx equals fraction numerator du over denominator 2 straight x minus 1 end fraction  integral left parenthesis 2 straight x minus 1 right parenthesis square root of straight x squared minus straight x plus 5 end root space dx  equals integral up diagonal strike left parenthesis 2 straight x minus 1 right parenthesis end strike space left parenthesis straight u right parenthesis to the power of 1 half end exponent fraction numerator du over denominator up diagonal strike left parenthesis 2 straight x minus 1 right parenthesis end strike end fraction  equals integral straight u to the power of 1 half end exponent du  equals 2 over 3 straight u to the power of 2 over 3 end exponent plus straight C  equals 2 over 3 left parenthesis straight x squared minus straight x plus 5 right parenthesis square root of straight x squared minus straight x plus 5 end root space plus straight C

0

Roboguru

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