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begin mathsize 14px style integral fraction numerator 2 x cubed minus 3 x squared plus 1 over denominator x squared end fraction d x equals horizontal ellipsis end style

  1. begin mathsize 14px style x squared minus 3 x plus 1 over x plus c end style

  2. begin mathsize 14px style x squared minus 3 x minus 1 over x plus c end style

  3. begin mathsize 14px style x squared minus 3 x plus 1 over x squared plus c end style

  4. begin mathsize 14px style x squared minus 3 x minus 1 over x squared plus c end style

  5. begin mathsize 14px style 1 half x squared minus 3 x plus 1 over x squared plus c end style

Pembahasan Soal:

Ingatlah bahwa

undefined

dan

begin mathsize 14px style integral k space d x equals k x plus c end style.

Kemudian perhatikan bahwa

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 2 x cubed minus 3 x squared plus 1 over denominator x squared end fraction end cell equals cell fraction numerator 2 x cubed over denominator x squared end fraction minus fraction numerator 3 x squared over denominator x squared end fraction plus 1 over x squared end cell row blank equals cell 2 x minus 3 plus x to the power of negative 2 end exponent end cell end table end style

Akibatnya

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

Jadi, jawaban yang tepat adalah B.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Acfreelance

Terakhir diupdate 02 Mei 2021

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Pembahasan Soal:

Integral pada soal merupakan integral tak tentu yang didefinisikan sebagai integral f left parenthesis x right parenthesis d x equals F left parenthesis x right parenthesis plus C. fungsi fraction numerator 2 x cubed minus 3 x squared plus 1 over denominator x squared end fraction dapat diubah menjadi fraction numerator 2 x cubed minus 3 x squared plus 1 over denominator x squared end fraction equals fraction numerator 2 x cubed over denominator x squared end fraction minus fraction numerator 3 x squared over denominator x squared end fraction plus 1 over x squared equals 2 x minus 3 plus x to the power of negative 2 end exponent comma sehingga diperoleh
 

integral fraction numerator 2 x cubed minus 3 x squared plus 1 over denominator x squared end fraction text d end text x equals integral 2 x minus 3 plus x to the power of negative 2 end exponent text d end text x


Berdasarkan sifat integral tak tentu yaitu integral x to the power of n d x equals fraction numerator x to the power of n plus 1 end exponent over denominator n plus 1 end fraction plus C comma n not equal to 1 comma maka
 

table attributes columnalign right center left columnspacing 0px end attributes row cell integral fraction numerator 2 x cubed minus 3 x squared plus 1 over denominator x squared end fraction text d end text x end cell equals cell integral 2 x minus 3 plus x to the power of negative 2 end exponent text d end text x end cell row cell integral fraction numerator 2 x cubed minus 3 x squared plus 1 over denominator x squared end fraction text d end text x end cell equals cell fraction numerator 2 x to the power of 1 plus 1 end exponent over denominator 1 plus 1 end fraction minus fraction numerator 3 x to the power of 0 plus 1 end exponent over denominator 0 plus 1 end fraction plus fraction numerator x to the power of negative 2 plus 1 end exponent over denominator negative 2 plus 1 end fraction plus C right parenthesis end cell row cell integral fraction numerator 2 x cubed minus 3 x squared plus 1 over denominator x squared end fraction d x end cell equals cell fraction numerator 2 x squared over denominator 2 end fraction minus fraction numerator 3 x over denominator 1 end fraction plus fraction numerator x to the power of negative 1 end exponent over denominator negative 1 end fraction plus C end cell row cell integral fraction numerator 2 x cubed minus 3 x squared plus 1 over denominator x squared end fraction d x end cell equals cell x squared minus 3 x minus 1 over x plus C end cell end table


Oleh karena itu, jawaban yang benar adalah B.

Roboguru

...

Pembahasan Soal:

Operasikan bentuk integral tak tentu tersebut seperti berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell integral fraction numerator 2 x cubed minus 3 x squared plus 1 over denominator x squared end fraction d x end cell equals cell integral fraction numerator 2 x cubed over denominator x squared end fraction minus fraction numerator 3 x squared over denominator x squared end fraction plus 1 over x squared d x end cell row blank equals cell integral 2 x minus 3 plus 1 over x squared d x end cell row blank equals cell integral 2 x d x minus integral 3 d x plus integral 1 over x squared d x end cell row blank equals cell integral 2 x d x minus integral 3 d x plus integral x to the power of negative 2 end exponent d x end cell row blank equals cell fraction numerator 2 over denominator 1 plus 1 end fraction x to the power of 1 plus 1 end exponent minus 3 over 1 x to the power of 0 plus 1 end exponent plus fraction numerator 1 over denominator negative 2 plus 1 end fraction x to the power of negative 2 plus 1 end exponent plus C end cell row blank equals cell 2 over 2 x squared minus 3 x to the power of 1 minus x to the power of negative 1 end exponent plus C end cell row blank equals cell x squared minus 3 x minus 1 over x plus C end cell end table

 

Maka, integral fraction numerator 2 x cubed minus 3 x squared plus 1 over denominator x squared end fraction d x equalsx squared minus 3 x minus 1 over x plus C

Jadi, jawaban yang tepat adalah B.

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Hasil dari  adalah ....

Pembahasan Soal:

Integral fungsi pada soal dapat dikerjakan dengan metode substitusi. Misalkan begin mathsize 14px style x squared minus 2 x plus 10 equals p end style  maka dengan menurunkan kedua ruas diperoleh begin mathsize 14px style left parenthesis 2 x minus 2 right parenthesis d x equals d p end style sehingga begin mathsize 14px style d x equals fraction numerator 1 over denominator 2 x minus 2 end fraction d p end style.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript blank fraction numerator x minus 1 over denominator square root of x squared minus 2 x plus 10 end root end fraction d x end cell equals cell integral subscript blank fraction numerator x minus 1 over denominator square root of p end fraction times fraction numerator 1 over denominator 2 x minus 2 end fraction d p end cell row blank equals cell 1 half integral fraction numerator 1 over denominator square root of p end fraction d p end cell row blank equals cell 1 half times 2 square root of p plus C end cell row blank equals cell square root of p plus C end cell end table end style            

dengan begin mathsize 14px style C end style konstanta. Selanjutnya dengan mensubstitusi kembali nilai undefined, diperoleh 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript blank fraction numerator x minus 1 over denominator square root of x squared minus 2 x plus 10 end root end fraction d x end cell equals cell square root of p plus C end cell row blank equals cell square root of x squared minus 2 x plus 10 end root plus C end cell end table end style .


Jadi, jawaban yang benar adalah D.

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Hasil dari  adalah ....

Pembahasan Soal:

Integral fungsi pada soal dapat dikerjakan dengan metode substitusi. Misalkan begin mathsize 14px style 1 plus 2 x minus x squared equals p end style  maka dengan menurunkan kedua ruas diperoleh begin mathsize 14px style left parenthesis 2 minus 2 x right parenthesis d x equals d p end style sehingga begin mathsize 14px style d x equals fraction numerator 1 over denominator 2 minus 2 x end fraction d p end style.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript blank fraction numerator x minus 1 over denominator left parenthesis 1 plus 2 x minus x squared right parenthesis cubed end fraction d x end cell equals cell integral subscript blank fraction numerator x minus 1 over denominator p cubed end fraction times fraction numerator 1 over denominator 2 minus 2 x end fraction d p end cell row blank equals cell negative integral subscript blank 1 over p cubed d p end cell row blank equals cell negative integral subscript blank p to the power of negative 3 end exponent d p end cell row blank equals cell negative fraction numerator 1 over denominator negative 3 plus 1 end fraction p to the power of negative 3 plus 1 end exponent plus C end cell row blank equals cell 1 half times 1 over p squared plus C end cell end table end style            

dengan begin mathsize 14px style C end style konstanta. Selanjutnya dengan mensubstitusi kembali nilai undefined, diperoleh 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript blank fraction numerator x minus 1 over denominator left parenthesis 1 plus 2 x minus x squared right parenthesis cubed end fraction d x end cell equals cell 1 half times 1 over p squared plus C end cell row blank equals cell fraction numerator 1 over denominator 2 left parenthesis 1 plus 2 x minus x squared right parenthesis squared end fraction plus C end cell end table end style .


Jadi, jawaban yang benar adalah B.

Roboguru

Hasil dari  adalah ....

Pembahasan Soal:

Perhatikan perhitungan berikut ini! 

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

Jika dimisalkan begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell C subscript 1 plus left parenthesis square root of 2 minus 2 right parenthesis C subscript 2 minus left parenthesis 2 square root of 2 plus 1 right parenthesis C subscript 3 plus square root of 2 C subscript 4 end cell end table 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 C end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank comma end table end style maka

size 14px 1 over size 14px 4 size 14px x to the power of size 14px 4 size 14px plus fraction numerator square root of size 14px 2 size 14px minus size 14px 1 over denominator size 14px 3 end fraction size 14px x to the power of size 14px 3 size 14px minus fraction numerator size 14px 2 square root of size 14px 2 size 14px plus size 14px 1 over denominator size 14px 2 end fraction size 14px x to the power of size 14px 2 size 14px plus square root of size 14px 2 size 14px x size 14px plus size 14px C subscript size 14px 1 size 14px plus size 14px left parenthesis square root of size 14px 2 size 14px minus size 14px 2 size 14px right parenthesis size 14px C subscript size 14px 2 size 14px minus begin mathsize 14px style left parenthesis 2 square root of 2 plus 1 right parenthesis end style size 14px C subscript size 14px 3 size 14px plus square root of size 14px 2 size 14px C subscript size 14px 4 size 14px equals size 14px 1 over size 14px 4 size 14px x to the power of size 14px 4 size 14px plus fraction numerator square root of size 14px 2 size 14px minus size 14px 1 over denominator size 14px 3 end fraction size 14px x to the power of size 14px 3 size 14px minus fraction numerator size 14px 2 square root of size 14px 2 size 14px plus size 14px 1 over denominator size 14px 2 end fraction size 14px x to the power of size 14px 2 size 14px plus square root of size 14px 2 size 14px x size 14px plus size 14px C 

Jadi, jawaban yang tepat adalah A.

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