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Hasil dari integral fraction numerator 2 x minus 1 over denominator open parentheses 2 x squared minus 2 x plus 7 close parentheses to the power of 9 end fraction d x equals....

Pembahasan Soal:

Gunakan konsep integral substitusi.

Akan ditentukan hasil dari integral fraction numerator 2 x minus 1 over denominator open parentheses 2 x squared minus 2 x plus 7 close parentheses to the power of 9 end fraction d x.

Misalkan u equals 2 x squared minus 2 x plus 7, sehingga turunannya adalah 

table attributes columnalign right center left columnspacing 2px end attributes row cell fraction numerator straight d u over denominator straight d x end fraction end cell equals cell 4 x minus 2 end cell row cell straight d u end cell equals cell open parentheses 4 x minus 2 close parentheses straight d x end cell row cell 1 half straight d u end cell equals cell open parentheses 2 x minus 1 close parentheses straight d x end cell end table

Perhatikan perhitungan berikut.

table attributes columnalign right center left columnspacing 2px end attributes row cell integral fraction numerator 2 x minus 1 over denominator open parentheses 2 x squared minus 2 x plus 7 close parentheses to the power of 9 end fraction d x end cell equals cell integral fraction numerator open parentheses 2 x minus 1 close parentheses straight d x over denominator open parentheses 2 x squared minus 2 x plus 7 close parentheses to the power of 9 end fraction end cell row blank equals cell integral fraction numerator begin display style 1 half end style straight d u over denominator open parentheses u close parentheses to the power of 9 end fraction end cell row blank equals cell integral 1 half u to the power of negative 9 end exponent straight d u end cell row blank equals cell 1 half integral u to the power of negative 9 end exponent straight d u end cell row blank equals cell 1 half times fraction numerator 1 over denominator negative 9 plus 1 end fraction u to the power of negative 9 plus 1 end exponent plus c end cell row blank equals cell 1 half times fraction numerator 1 over denominator negative 8 end fraction u to the power of negative 8 end exponent plus c end cell row blank equals cell negative 1 over 16 u to the power of negative 8 end exponent plus c end cell row blank equals cell negative fraction numerator 1 over denominator 16 times u to the power of 8 end fraction plus c end cell row cell integral fraction numerator 2 x minus 1 over denominator open parentheses 2 x squared minus 2 x plus 7 close parentheses to the power of 9 end fraction d x end cell equals cell negative fraction numerator 1 over denominator 16 times open parentheses 2 x squared minus 2 x plus 7 close parentheses to the power of 8 end fraction plus c end cell end table

Jadi, diperoleh hasil dari integral fraction numerator 2 x minus 1 over denominator open parentheses 2 x squared minus 2 x plus 7 close parentheses to the power of 9 end fraction d x equals negative fraction numerator 1 over denominator 16 times open parentheses 2 x squared minus 2 x plus 7 close parentheses to the power of 8 end fraction plus c.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Dwi

Mahasiswa/Alumni Universitas Negeri Yogyakarta

Terakhir diupdate 04 Juni 2021

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Pertanyaan yang serupa

Hasil dari = ...

Pembahasan Soal:

Ingat konsep turunan dan integral substitusi:

table row cell u equals a x to the power of n end cell rightwards arrow cell fraction numerator straight d u over denominator straight d x end fraction equals a times n x to the power of n minus 1 end exponent end cell row cell integral a x to the power of n end cell rightwards arrow cell fraction numerator a over denominator n plus 1 end fraction x to the power of n plus 1 end exponent plus C end cell end table 

Misalkan u equals x squared minus 4 x plus 1 maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator straight d u over denominator straight d x end fraction end cell equals cell 2 x minus 4 end cell row dx equals cell fraction numerator 1 over denominator 2 x minus 4 end fraction straight d u end cell end table 

kemudian substitusikan:

table attributes columnalign right center left columnspacing 0px end attributes row cell integral open parentheses x minus 2 close parentheses open parentheses x squared minus 4 x plus 1 close parentheses cubed space straight d x end cell equals cell integral open parentheses x minus 2 close parentheses u cubed space fraction numerator 1 over denominator 2 x minus 4 end fraction space straight d u end cell row blank equals cell integral fraction numerator x minus 2 over denominator 2 open parentheses x minus 2 close parentheses end fraction u cubed space straight d u end cell row blank equals cell integral 1 half u cubed space straight d u end cell row blank equals cell 1 half integral u squared space straight d u end cell row blank equals cell 1 half open square brackets 1 third u to the power of 4 plus C close square brackets end cell row blank equals cell 1 over 6 open parentheses x squared minus 4 x plus 1 close parentheses to the power of 4 plus C end cell end table 

Dengan demikian hasil dari integral open parentheses x minus 2 close parentheses open parentheses x squared minus 4 x plus 1 close parentheses cubed space straight d x adalah Error converting from MathML to accessible text..

0

Roboguru

Hasil dari

Pembahasan Soal:

Menggunakan metode subtitusi, akan ditentukan hasil dari

 integral 2 x square root of x squared minus 2 end root d x

Misalkan u equals x squared minus 2. Berarti

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

Maka dapat diperoleh

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral 2 x square root of x squared minus 2 end root d x end cell equals cell integral open parentheses square root of x squared minus 2 end root close parentheses 2 x d x end cell row blank equals cell integral square root of u d u end cell row blank equals cell integral u to the power of 1 half end exponent d u end cell row blank equals cell 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 comma space untuk space C space suatu space konstanta end cell row blank equals cell fraction numerator 1 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 2 over 3 u square root of u plus C end cell row blank equals cell 2 over 3 open parentheses x squared minus 2 close parentheses square root of x squared minus 2 end root plus C end cell end table end style 

Jadi, jawaban yang tepat adalah E.space  

0

Roboguru

Hasil dari  adalah

Pembahasan Soal:

Kita akan menyelesaiakan permasalahan integral tersebut menggunakan metode substitusi.

Misalkan begin mathsize 14px style u equals 2 x squared minus 6 x plus 7 end style, maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator d u over denominator d x end fraction end cell equals cell 4 x minus 6 end cell row cell d x end cell equals cell fraction numerator d u over denominator 4 x minus 6 end fraction end cell end table end style

Akibatnya

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell integral open parentheses 2 x minus 3 close parentheses square root of 2 x squared minus 6 x plus 7 end root d x end cell row blank equals cell integral open parentheses 2 x minus 3 close parentheses square root of u fraction numerator d u over denominator 4 x minus 6 end fraction end cell row blank equals cell integral up diagonal strike open parentheses 2 x minus 3 close parentheses end strike u to the power of 1 half end exponent fraction numerator d u over denominator 2 up diagonal strike open parentheses 2 x minus 3 close parentheses end strike end fraction end cell row blank equals cell 1 half integral u to the power of 1 half end exponent d u end cell row blank equals cell fraction numerator 1 over denominator up diagonal strike 2 end fraction open parentheses fraction numerator up diagonal strike 2 over denominator 3 end fraction u to the power of 3 over 2 end exponent close parentheses plus c end cell row blank equals cell 1 third u square root of u plus c end cell row blank equals cell 1 third open parentheses 2 x squared minus 6 x plus 7 close parentheses square root of 2 x squared minus 6 x plus 7 end root plus c end cell end table end style

Jadi, jawaban yang tepat adalah C.

0

Roboguru

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell integral 8 square root of open parentheses x squared minus 2 x to the power of 4 close parentheses end root d x end cell equals cell integral 8 square root of x squared open parentheses 1 minus 2 x squared close parentheses end root d x end cell row blank equals cell integral 8 x square root of open parentheses 1 minus 2 x squared close parentheses end root d x end cell row blank equals cell integral 8 x times open parentheses 1 minus 2 x squared close parentheses to the power of 1 half end exponent d x end cell end table

Misalkan:

table attributes columnalign right center left columnspacing 0px end attributes row u equals cell open parentheses 1 minus 2 x squared close parentheses end cell row cell d u end cell equals cell negative 4 x space d x end cell row cell d x end cell equals cell fraction numerator d u over denominator negative 4 x end fraction end cell end table 

Sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell integral 8 x times open parentheses 1 minus 2 x squared close parentheses to the power of 1 half end exponent d x end cell equals cell integral 8 x times u blank to the power of 1 half end exponent fraction numerator d u over denominator negative 4 x end fraction end cell row blank equals cell integral negative 2 times u to the power of 1 half end exponent d u end cell row blank equals cell negative 4 over 3 u to the power of 3 over 2 end exponent plus C end cell row blank equals cell negative 4 over 3 open parentheses 1 minus 2 x squared close parentheses to the power of 3 over 2 end exponent plus C end cell end table

Jadi hasil dari integral 8 square root of open parentheses x squared minus 2 x to the power of 4 close parentheses end root d x adalah Error converting from MathML to accessible text.

 

0

Roboguru

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    

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