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Selesaikan integral dengan substitusi berikut:

Pertanyaan

Selesaikan integral dengan substitusi berikut:
begin mathsize 14px style straight a. space integral open parentheses fraction numerator 3 x d x over denominator x squared plus 2 end fraction close parentheses subscript blank end style 

Pembahasan Soal:

Dengan integral subtitusi didapat:

begin mathsize 14px style table row cell Misal colon end cell u equals cell x squared plus 2 end cell blank blank blank blank row blank cell d u end cell equals cell 2 x end cell cell d x end cell blank blank blank row blank cell d x end cell equals cell fraction numerator d u over denominator 2 x end fraction end cell blank blank blank blank row cell integral subscript blank open parentheses fraction numerator 3 x over denominator x squared plus 2 end fraction close parentheses d x end cell equals cell integral subscript blank open parentheses fraction numerator 3 x over denominator u end fraction close parentheses fraction numerator d u over denominator 2 x end fraction end cell blank blank blank blank blank row blank equals cell 3 over 2 integral subscript blank subscript blank fraction numerator d u over denominator u end fraction end cell blank blank blank blank blank row blank equals cell 3 over 2 ln space vertical line u vertical line plus c end cell blank blank blank blank blank row blank equals cell 3 over 2 ln space vertical line x squared plus 2 vertical line plus c end cell blank blank blank blank blank end table end style 

Dengan demikian, penyelesaian dari soal di atas adalah 23lnx2+2∣+c 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Acfreelance

Mahasiswa/Alumni Universitas Siliwangi

Terakhir diupdate 12 September 2021

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Pertanyaan yang serupa

Tentukan hasil integral fungsi aljabar berikut.

Pembahasan Soal:

Integral fungsi pada soal dapat dikerjakan dengan metode substitusi. Misalkan begin mathsize 14px style x squared minus 3 equals p end style maka dengan menurunkan kedua ruas diperoleh begin mathsize 14px style 2 x d x equals d p end style sehingga begin mathsize 14px style d x equals fraction numerator 1 over denominator 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 x left parenthesis x squared minus 3 right parenthesis cubed d x end cell equals cell integral subscript blank x times p cubed times fraction numerator 1 over denominator 2 x end fraction d p end cell row blank equals cell 1 half integral subscript blank p cubed d p end cell row blank equals cell 1 half times fraction numerator 1 over denominator 3 plus 1 end fraction p to the power of 3 plus 1 end exponent plus C end cell row blank equals cell 1 over 8 p to the power of 4 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 x left parenthesis x squared minus 3 right parenthesis cubed d x end cell equals cell fraction numerator 1 over denominator 8 end fraction p to the power of 4 plus C end cell row blank equals cell 1 over 8 open parentheses x squared minus 3 close parentheses to the power of 4 plus C end cell end table end style    .


Sehingga

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript negative 1 end subscript superscript 2 x left parenthesis x squared minus 3 right parenthesis cubed d x end cell equals cell 1 over 8 left parenthesis x squared minus 3 right parenthesis to the power of 4 left enclose blank with negative 1 below and 2 on top end enclose space end cell row blank equals cell open square brackets 1 over 8 left parenthesis 2 squared minus 3 right parenthesis to the power of 4 close square brackets minus open square brackets 1 over 8 left parenthesis left parenthesis negative 1 right parenthesis squared minus 3 right parenthesis to the power of 4 close square brackets end cell row blank equals cell 1 over 8 minus 16 over 8 end cell row blank equals cell negative 15 over 8 end cell end table end style  

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Roboguru

Hasil dari

Pembahasan Soal:

integral subscript space superscript space fraction numerator 2 x over denominator square root of x squared plus 1 end root end fraction space d x space equals integral subscript space superscript space fraction numerator 1 over denominator square root of x squared plus 1 end root end fraction space d left parenthesis x squared plus 1 right parenthesis space space space left square bracket h a l space i n i space d i k a r e n a k a n space d left parenthesis x squared plus 1 right parenthesis equals 2 x space d x right square bracket  equals integral subscript space superscript space open parentheses x squared plus 1 close parentheses to the power of negative 1 half end exponent space space d left parenthesis x squared plus 1 right parenthesis space  equals fraction numerator 1 over denominator negative 1 half plus 1 end fraction open parentheses x squared plus 1 close parentheses to the power of negative 1 half plus 1 end exponent plus C  equals 2 open parentheses x squared plus 1 close parentheses to the power of 1 half end exponent plus C  equals 2 square root of x squared plus 1 end root plus C

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Roboguru

Pembahasan Soal:

Menggunakan metode subtitusi, akan dicari nilai integral di atas. Misalkan u equals x cubed minus 5. Maka didapat

table attributes columnalign right center left columnspacing 0px end attributes row u equals cell x cubed minus 5 end cell row cell fraction numerator d u over denominator d x end fraction end cell equals cell fraction numerator d open parentheses x cubed minus 5 close parentheses over denominator d x end fraction end cell row cell fraction numerator d u over denominator d x end fraction end cell equals cell 3 x squared end cell row cell d u end cell equals cell 3 x squared d x end cell row cell 2 cross times d u end cell equals cell 2 cross times 3 x squared d x end cell row cell 2 d u end cell equals cell 6 x squared d x end cell end table

Dengan demikian diperoleh

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

Jadi, integral 6 x squared square root of x cubed minus 5 end root d x equals 4 over 3 open parentheses x cubed minus 5 close parentheses square root of x cubed minus 5 end root plus C 

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Roboguru

Tentukan hasil dari_ iritegrasi berikut. a.

Pembahasan Soal:

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Roboguru

Hasil dari

Pembahasan Soal:

M i s a l colon space u equals 2 x squared minus 6 x plus 5  d u equals 4 x minus 6 space d x  d u equals 2 left parenthesis 2 x minus 3 right parenthesis d x  d x equals fraction numerator d u over denominator 2 left parenthesis 2 x minus 3 right parenthesis end fraction    integral fraction numerator left parenthesis 2 x minus 3 right parenthesis d x over denominator square root of 2 x squared minus 6 x plus 5 end root end fraction equals integral fraction numerator begin display style fraction numerator 2 x minus 3 right parenthesis d u over denominator 2 left parenthesis 2 x minus 3 right parenthesis end fraction end style over denominator u to the power of begin display style 1 half end style end exponent end fraction equals 1 half integral fraction numerator d u over denominator u to the power of begin display style 1 half end style end exponent end fraction equals 1 half integral u to the power of fraction numerator negative 1 over denominator 2 end fraction end exponent d u    equals 1 half. fraction numerator 1 over denominator begin display style 1 half end style end fraction u to the power of 1 half end exponent equals 1 half.2. u to the power of 1 half end exponent equals u to the power of 1 half end exponent equals square root of 2 x squared minus 6 x plus 5 end root plus C

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