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Pertanyaan

integral square root of 4 x cubed plus 2 x squared end root space d x equals horizontal ellipsis

Pembahasan Soal:

Diberikan bentuk aljabar seperti pada soal. Fungsi di dalamnya dapat disederhanakan seperti berikut :

table attributes columnalign right center left columnspacing 0px end attributes row cell integral square root of 4 x cubed plus 2 x squared end root space d x end cell equals cell integral square root of x squared times open parentheses 4 x plus 2 close parentheses end root space d x end cell row blank equals cell integral x square root of 4 x plus 2 end root space d x end cell end table 

Dimisalkan :

u equals x space rightwards arrow space d u equals d x

 table attributes columnalign right center left columnspacing 0px end attributes row cell d v end cell equals cell square root of 4 x plus 2 end root space d x end cell row cell integral d v end cell equals cell integral left parenthesis 4 x plus 2 right parenthesis to the power of 1 half end exponent space d x end cell row v equals cell 1 fourth times 2 over 3 times left parenthesis 4 x plus 2 right parenthesis to the power of 3 over 2 end exponent plus C end cell row v equals cell 1 over 6 left parenthesis 4 x plus 2 right parenthesis to the power of 3 over 2 end exponent plus C end cell end table

Dengan demikian, penyelesaian bentuk integral di atas adalah

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell integral x square root of 4 x plus 2 end root space d x end cell row blank equals cell u v minus integral v space d u end cell row blank equals cell x open parentheses 1 over 6 left parenthesis 4 x plus 2 right parenthesis to the power of 3 over 2 end exponent close parentheses minus 1 over 6 integral left parenthesis 4 x plus 2 right parenthesis to the power of 3 over 2 end exponent space d x end cell row blank equals cell x over 6 left parenthesis 4 x plus 2 right parenthesis to the power of 3 over 2 end exponent minus 1 over 6 times 1 fourth times 2 over 5 left parenthesis 4 x plus 2 right parenthesis to the power of 5 over 2 end exponent plus C end cell row blank equals cell 1 over 6 left parenthesis 4 x plus 2 right parenthesis to the power of 3 over 2 end exponent open parentheses x minus 1 over 10 left parenthesis 4 x plus 2 right parenthesis close parentheses plus C end cell row blank equals cell 1 over 6 left parenthesis 4 x plus 2 right parenthesis to the power of 3 over 2 end exponent open parentheses fraction numerator 10 x minus 4 x minus 2 over denominator 10 end fraction close parentheses plus C end cell row blank equals cell 1 over 6 left parenthesis 4 x plus 2 right parenthesis to the power of 3 over 2 end exponent open parentheses fraction numerator 6 x minus 2 over denominator 10 end fraction close parentheses plus C end cell row blank equals cell 1 over 6 times 2 over 10 left parenthesis 4 x plus 2 right parenthesis to the power of 3 over 2 end exponent left parenthesis 3 x minus 1 right parenthesis plus C end cell row blank equals cell 1 over 30 left parenthesis 4 x plus 2 right parenthesis to the power of 3 over 2 end exponent left parenthesis 3 x minus 1 right parenthesis plus C end cell end table  

Jadi, hasil penyelesaiannya adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 1 over 30 left parenthesis 4 x plus 2 right parenthesis to the power of 3 over 2 end exponent left parenthesis 3 x minus 1 right parenthesis plus C end cell end table.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Kumaralalita

Mahasiswa/Alumni Universitas Gadjah Mada

Terakhir diupdate 06 Juni 2021

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Pertanyaan yang serupa

Pembahasan Soal:

Untuk menyelesaikan integral tersebut, dapat menggunakan metode substitusi.

Misalkan,

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row u equals cell x plus 1 end cell row cell fraction numerator d u over denominator d x end fraction end cell equals 1 row cell d x end cell equals cell d u end cell row blank blank blank row u equals cell x plus 1 rightwards arrow x equals u minus 1 end cell end table end style

Kemudian substitusikan, sehingga didapat:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral fraction numerator x over denominator x plus 1 end fraction d x end cell equals cell integral fraction numerator u minus 1 over denominator u end fraction d u end cell row blank equals cell integral 1 minus 1 over u d u end cell row blank equals cell u minus ln open vertical bar u close vertical bar plus C end cell end table end style  

Substitusikan kembali begin mathsize 14px style u equals x plus 1 end style.

Jadi, hasilnya adalah begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank integral end table fraction numerator x over denominator x plus 1 end fraction d x table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank x 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 1 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank ln end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell vertical line x plus 1 vertical line 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 end style.

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Roboguru

Hasil dari

Pembahasan Soal:

integral open parentheses 3 straight x plus 1 close parentheses square root of 3 straight x squared plus 2 straight x minus 4 end root straight space dx  Gunakan space integral space subtitusi  straight u equals 3 straight x squared plus 2 straight x minus 4  du equals 6 straight x plus 2 space dx  1 half du equals 3 straight x plus 1 space dx  Maka space bentuk space pada space soal space dapat space ditulis space colon  integral open parentheses 3 straight x plus 1 close parentheses square root of 3 straight x squared plus 2 straight x minus 4 end root straight space dx  equals integral square root of 3 straight x squared plus 2 straight x minus 4 end root open parentheses 3 straight x plus 1 close parentheses straight space dx  equals integral square root of straight u open parentheses 1 half du close parentheses  equals 1 half integral straight u to the power of 1 half end exponent straight space du  equals 1 half.2 over 3 straight u to the power of 3 over 2 end exponent plus straight C  equals 1 third open parentheses 3 straight x squared plus 2 straight x minus 4 close parentheses to the power of 3 over 2 end exponent plus straight C

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Roboguru

Hasil dari

Pembahasan Soal:

integral cube root of open parentheses 2 x minus 1 close parentheses squared end root d x equals integral left parenthesis 2 x minus 1 right parenthesis to the power of 2 over 3 end exponent d x  equals 3 over 5 open parentheses 2 x minus 1 close parentheses to the power of 5 over 3 end exponent 1 half plus c  equals 3 over 10 left parenthesis 2 x minus 1 right parenthesis to the power of 5 over 3 end exponent plus c

equals 3 over 10 left parenthesis x minus 1 right parenthesis cube root of left parenthesis 2 x minus 1 right parenthesis squared end root plus c

0

Roboguru

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 

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Roboguru

Hasil dari

Pembahasan Soal:

begin mathsize 14px style integral open parentheses 3 x plus 2 close parentheses open parentheses 3 x squared plus 4 x minus 1 close parentheses squared d x  misalkan colon space u equals 3 x squared plus 4 x minus 1 fraction numerator d u over denominator d x end fraction equals 6 x plus 4 1 half fraction numerator d u over denominator d x end fraction equals 3 x plus 2 1 half straight d u equals 3 x plus 2 space straight d x  integral open parentheses 3 x plus 2 close parentheses open parentheses 3 x squared plus 4 x minus 1 close parentheses squared d x equals integral 1 half space u squared space straight d u equals 1 half times fraction numerator 1 over denominator 2 plus 1 end fraction u cubed plus straight C equals 1 over 6 u cubed plus straight C equals 1 over 6 left parenthesis 3 x squared plus 4 x minus 1 right parenthesis cubed plus straight C space space end style  

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