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∫36x−4​4​

Pertanyaan

integral fraction numerator 4 over denominator cube root of 6 x minus 4 end root end fraction 

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral fraction numerator 4 over denominator cube root of 6 x minus 4 end root end fraction d x end cell equals cell integral 4 open parentheses 6 x minus 4 close parentheses to the power of negative 1 third end exponent d x end cell row blank equals cell 4 times fraction numerator 1 over denominator negative begin display style 1 third end style plus 1 end fraction times 1 over 6 open parentheses 6 x minus 4 close parentheses to the power of negative 1 third plus 1 end exponent plus c end cell row blank equals cell 4 times fraction numerator 1 over denominator begin display style 2 over 3 end style end fraction times 1 over 6 open parentheses 6 x minus 4 close parentheses to the power of 2 over 3 end exponent plus c end cell row blank equals cell 4 times 3 over 2 times 1 over 6 open parentheses 6 x minus 4 close parentheses 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 4 close parentheses squared end root plus c end cell end table end style    

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Mustikowati

Mahasiswa/Alumni Universitas Negeri Jakarta

Terakhir diupdate 06 Oktober 2021

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Pertanyaan yang serupa

∫01​(x2+2x+6)x+1​dx=...

Pembahasan Soal:

Permasalahan integral di atas akan diselesaikan dengan metode integral subtitusi. 

Misalkan u equals x squared plus 2 x plus 6, maka diperoleh:

  • fraction numerator d u over denominator d x end fraction equals 2 x plus 2 equals 2 open parentheses x plus 1 close parentheses sehingga fraction numerator d u over denominator 2 end fraction equals open parentheses x plus 1 close parentheses d x 
  • batas bawah x equals 0 menjadi u equals 0 squared plus 2 open parentheses 0 close parentheses plus 6 equals 0 plus 0 plus 6 equals 6
  • batas atas x equals 1 menjadi u equals 1 squared plus 2 open parentheses 1 close parentheses plus 6 equals 1 plus 2 plus 6 equals 9

Dengan menyubtitusikan variable ufraction numerator d u over denominator 2 end fraction, batas bawah dan batas akhir di atas ke bentuk integral pada soal, diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript 0 superscript 1 fraction numerator x plus 1 over denominator open parentheses x squared plus 2 x plus 6 close parentheses end fraction d x end cell equals cell integral subscript 0 superscript 1 fraction numerator 1 over denominator open parentheses x squared plus 2 x plus 6 close parentheses end fraction open parentheses x plus 1 close parentheses d x end cell row blank equals cell integral subscript 6 superscript 9 1 over u fraction numerator d u over denominator 2 end fraction end cell row blank equals cell integral subscript 6 superscript 9 fraction numerator 1 over denominator 2 u end fraction d u end cell end table

Oleh karenaintegral subscript a superscript b 1 over x d x equals right enclose ln open parentheses x close parentheses end enclose subscript a superscript b equals ln open parentheses b close parentheses minus ln open parentheses a close parentheses untuk x greater than 0, maka penyelesaian integral di atas adalah 

 table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript 6 superscript 9 fraction numerator 1 over denominator 2 u end fraction d u end cell equals cell 1 half open parentheses integral subscript 6 superscript 9 1 over u d u close parentheses end cell row blank equals cell 1 half open parentheses right enclose ln open parentheses u close parentheses end enclose subscript 6 superscript 9 close parentheses end cell row blank equals cell 1 half open parentheses ln open parentheses 9 close parentheses minus ln open parentheses 6 close parentheses close parentheses end cell end table

Jadi, integral subscript 0 superscript 1 fraction numerator x plus 1 over denominator open parentheses x squared plus 2 x plus 6 close parentheses end fraction d x equals 1 half open parentheses ln open parentheses 9 close parentheses minus ln open parentheses 6 close parentheses close parentheses.space 

0

Roboguru

Hasil dari ∫(x−2)3dx=...

Pembahasan Soal:

integral open parentheses x minus 2 close parentheses cubed d x equals 1 fourth left parenthesis x minus 2 right parenthesis to the power of 4 plus c

0

Roboguru

∫x(x2−1)3dx=…

Pembahasan Soal:

Misalkan u equals x squared minus 1, maka didapatkan:

table attributes columnalign right center left columnspacing 0px end attributes row cell straight d u end cell equals cell 2 x space straight d x end cell row cell straight d x end cell equals cell fraction numerator straight d u over denominator 2 x end fraction end cell end table 

Substitusikan ke integral x open parentheses x squared minus 1 close parentheses cubed space straight d x dan didapatkan:

table attributes columnalign right center left columnspacing 0px end attributes row cell integral x open parentheses x squared minus 1 close parentheses cubed space straight d x end cell equals cell integral x times u cubed space fraction numerator straight d u over denominator 2 x end fraction end cell row blank equals cell integral 1 half u cubed space straight d u end cell row blank equals cell 1 half times 1 fourth u to the power of 4 plus c end cell row blank equals cell 1 over 8 u to the power of 4 plus c end cell row blank equals cell 1 over 8 open parentheses x squared minus 1 close parentheses to the power of 4 plus c end cell end table 

Jadi, integral x open parentheses x squared minus 1 close parentheses cubed space straight d x equals 1 over 8 open parentheses x squared minus 1 close parentheses to the power of 4 plus c.

0

Roboguru

Tentukan hasil dari integral berikut: f. Tentukan hasil dari ∫(18x2−12x)x3−x2+2​dx.

Pembahasan Soal:

Dengan menggunakan metode subtitusi

table attributes columnalign right center left columnspacing 0px end attributes row U equals cell x cubed minus x squared plus 2 end cell row cell fraction numerator d U over denominator d x end fraction end cell equals cell 3 x squared minus 2 x end cell row cell d x end cell equals cell fraction numerator d U over denominator 3 x squared minus 2 x end fraction end cell end table

Sehingga

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell integral open parentheses 18 x squared minus 12 x close parentheses square root of x cubed minus x squared plus 2 end root space d x end cell row blank equals cell integral 6 open parentheses 3 x squared minus 2 x close parentheses square root of U fraction numerator d U over denominator 3 x squared minus 2 x end fraction end cell row blank equals cell integral 6 square root of U d U end cell row blank equals cell integral 6 U to the power of 1 half end exponent d U end cell row blank equals cell fraction numerator 6 over denominator begin display style 1 half end style plus 1 end fraction U to the power of 1 half plus 1 end exponent plus C end cell row blank equals cell fraction numerator 6 over denominator begin display style 3 over 2 end style end fraction U to the power of 3 over 2 end exponent plus C end cell row blank equals cell 6 cross times 2 over 3 U square root of U plus C end cell row blank equals cell 4 U square root of U plus C end cell row blank equals cell 4 open parentheses x cubed minus x squared plus 2 close parentheses square root of x cubed minus x squared plus 2 end root plus C end cell end table

Dengan demikian hasil integral fungsi di atas adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank 4 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses x cubed minus x squared plus 2 close parentheses end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of x cubed minus x squared plus 2 end root end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank plus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank C end table.

0

Roboguru

Hasil dari

Pembahasan Soal:

d open parentheses x squared minus 2 x close parentheses equals open parentheses 2 x minus 2 close parentheses d x  d open parentheses x squared minus 2 x close parentheses equals 2 open parentheses x minus 1 close parentheses d x  open parentheses x minus 1 close parentheses d x equals 1 half d open parentheses x squared minus 2 x close parentheses    M a k a comma  integral fraction numerator open parentheses x minus 1 close parentheses over denominator square root of x squared minus 2 x end root end fraction d x equals integral fraction numerator 1 half over denominator open parentheses x squared minus 2 x close parentheses to the power of 1 half end exponent end fraction d open parentheses x squared minus 2 x close parentheses  space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1 half integral open parentheses x squared minus 2 x close parentheses to the power of negative space 1 half end exponent d open parentheses x squared minus 2 x close parentheses  space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1 half open parentheses 2 open parentheses x squared minus 2 x close parentheses to the power of 1 half end exponent close parentheses  space space space space space space space space space space space space space space space space space space space space space space space space space space equals square root of x squared minus 2 x end root plus C

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