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Pertanyaan

integral subscript 0 superscript 1 ln space x d x

Pembahasan Soal:

integral subscript 0 superscript 1 ln space x d x

 

table attributes columnalign right center left columnspacing 0px end attributes row cell misal comma space straight u space end cell equals cell space ln vertical line straight x vertical line space end cell row cell du space end cell equals cell space 1 divided by straight x space dx space space end cell row cell dv space end cell equals cell space dx space end cell row cell straight v space end cell equals cell space straight x space space end cell row blank equals cell space uv space minus space integral subscript 0 superscript 1 straight v space du space end cell row blank equals cell space ln vertical line straight x vertical line. straight x space minus space integral subscript 0 superscript 1 straight x space open parentheses 1 over straight x close parentheses space dx space end cell row blank equals cell space straight x. ln vertical line straight x vertical line space minus space integral subscript 0 superscript 1 dx space end cell row blank equals cell space straight x. ln vertical line straight x vertical line space minus space space large x subscript 0 superscript 1 plus space straight C space end cell row blank equals cell space space straight x. ln vertical line straight x vertical line minus 1 plus space straight C end cell end table

Jadi, integral subscript 0 superscript 1 ln space x d x adalah x space ln space open vertical bar x close vertical bar space minus 1 space plus C.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Rahmawati

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Terakhir diupdate 05 Juni 2021

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Pertanyaan yang serupa

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Pembahasan Soal:

Untuk memudahkan perhitungan integral, hitung terlebih dahulu hasil perkalian ketiga fungsi diatas.

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell integral open parentheses open parentheses x minus 1 over x close parentheses open parentheses x plus 1 over x close parentheses close parentheses open parentheses x squared minus x close parentheses space d x end cell row blank equals cell integral open parentheses x squared minus open parentheses 1 over x close parentheses squared close parentheses open parentheses x squared minus x close parentheses space d x end cell row blank equals cell integral open parentheses x squared minus 1 over x squared close parentheses open parentheses x squared minus x close parentheses space d x end cell row blank equals cell integral open parentheses x squared times x squared close parentheses minus open parentheses x squared times x close parentheses minus open parentheses 1 over x squared times x squared close parentheses minus open parentheses 1 over x squared times x close parentheses d x end cell row blank equals cell integral x to the power of 4 minus x cubed minus 1 minus 1 over x space d x end cell row blank equals cell fraction numerator 1 over denominator 4 plus 1 end fraction x to the power of 4 plus 1 end exponent minus fraction numerator 1 over denominator 3 plus 1 end fraction x to the power of 3 plus 1 end exponent minus x to the power of 0 plus 1 end exponent minus ln space x plus C end cell row blank equals cell 1 fifth x to the power of 5 minus 1 fourth x to the power of 4 minus x minus ln space x plus C end cell end table

Jadi, hasil dari integral di atas adalah

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 1 fifth x to the power of 5 minus 1 fourth x to the power of 4 minus x minus ln space x plus C end cell end table.

Roboguru

Diketahui  dan  adalah bilangan bulat positif. Jika , maka nilai dari adalah ....

Pembahasan Soal:

Dari soal, diketahui hasil integral sebagai berikut.

integral x to the power of n ln open parentheses x close parentheses space d x equals 1 over 16 x to the power of 4 open parentheses 4 ln open parentheses x close parentheses minus 1 close parentheses plus C

Terlebih dahulu, bentuk integral x to the power of n ln open parentheses x close parentheses akan diselesaikan dengan menerapkan metode integral parsial.

Misal u equals ln open parentheses x close parentheses dan fraction numerator d v over denominator d x end fraction equals x to the power of n, maka didapat perhitungan sebagai berikut.

fraction numerator d u over denominator d x end fraction equals 1 over x

dan

table attributes columnalign right center left columnspacing 0px end attributes row v equals cell integral fraction numerator d v over denominator d x end fraction space d x end cell row blank equals cell integral x to the power of n space d x end cell row blank equals cell fraction numerator x to the power of n plus 1 end exponent over denominator n plus 1 end fraction plus C end cell end table

Selanjutnya, perhatikan perhitungan berikut!

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell integral x to the power of n ln open parentheses x close parentheses space d x end cell row blank equals cell integral u fraction numerator d v over denominator d x end fraction space d x end cell row blank equals cell u v minus integral v fraction numerator d u over denominator d x end fraction space d x end cell row blank equals cell ln open parentheses x close parentheses times fraction numerator x to the power of n plus 1 end exponent over denominator n plus 1 end fraction minus integral fraction numerator x to the power of n plus 1 end exponent over denominator n plus 1 end fraction times 1 over x space d x end cell row blank equals cell ln open parentheses x close parentheses times fraction numerator x to the power of n plus 1 end exponent over denominator n plus 1 end fraction minus integral fraction numerator x to the power of n over denominator n plus 1 end fraction space d x end cell row blank equals cell ln open parentheses x close parentheses times fraction numerator x to the power of n plus 1 end exponent over denominator n plus 1 end fraction minus open parentheses fraction numerator 1 over denominator n plus 1 end fraction times fraction numerator 1 over denominator n plus 1 end fraction x to the power of n plus 1 end exponent close parentheses plus C end cell row blank equals cell ln open parentheses x close parentheses times fraction numerator x to the power of n plus 1 end exponent over denominator n plus 1 end fraction minus x to the power of n plus 1 end exponent over open parentheses n plus 1 close parentheses squared plus C end cell row blank equals cell 1 over open parentheses n plus 1 close parentheses squared x to the power of n plus 1 end exponent open parentheses open parentheses n plus 1 close parentheses times ln open parentheses x close parentheses minus 1 close parentheses plus C end cell end table

Diketahui bahwa integral x to the power of n ln open parentheses x close parentheses space d x equals 1 over 16 x to the power of 4 open parentheses 4 ln open parentheses x close parentheses minus 1 close parentheses plus C, maka didapat hubungan berikut ini.

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 1 over open parentheses n plus 1 close parentheses squared x to the power of n plus 1 end exponent open parentheses open parentheses n plus 1 close parentheses times ln open parentheses x close parentheses minus 1 close parentheses plus C end cell row blank equals cell 1 over 16 x to the power of 4 open parentheses 4 ln open parentheses x close parentheses minus 1 close parentheses plus C end cell end table

Dengan menyamakan suku pada persamaan di atas, didapat hasil sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses n plus 1 close parentheses ln invisible function application open parentheses x close parentheses end cell equals cell 4 ln invisible function application open parentheses x close parentheses end cell row cell n plus 1 end cell equals 4 row n equals 3 end table end style

Jika disubstitusi n equals 3 ke bagian yang lain, dapat diperiksa bahwa persamaan tersebut terpenuhi.

Dengan demikian, nilai dari n adalah 3.

Jadi, jawaban yang tepat adalah B.

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Carilah !

Pembahasan Soal:

Dengan menggunakan sifat integral, maka diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell integral fraction numerator 3 over denominator x squared minus 1 end fraction d x end cell equals cell 3 times open parentheses integral fraction numerator 1 over denominator x squared minus 1 end fraction d x close parentheses end cell row blank equals cell 3 times open parentheses integral fraction numerator 1 over denominator negative left parenthesis negative x squared plus 1 right parenthesis end fraction d x close parentheses end cell row blank equals cell 3 times open parentheses negative integral fraction numerator 1 over denominator negative x squared plus 1 end fraction d x close parentheses end cell row blank equals cell 3 times open parentheses negative open parentheses fraction numerator ln space open vertical bar x plus 1 close vertical bar over denominator 2 end fraction close parentheses minus fraction numerator ln space open vertical bar x minus 1 close vertical bar over denominator 2 end fraction close parentheses plus C end cell row cell integral fraction numerator 3 over denominator x squared minus 1 end fraction d x end cell equals cell negative 3 open parentheses 1 half ln space open vertical bar x plus 1 close vertical bar minus 1 half ln space open vertical bar x minus 1 close vertical bar close parentheses plus C end cell end table 

Dengan demikian, nilai dari integral fraction numerator 3 over denominator x squared minus 1 end fraction d x adalah 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 3 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses 1 half ln space open vertical bar x plus 1 close vertical bar minus 1 half ln space open vertical bar x minus 1 close vertical bar close parentheses 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 table attributes columnalign right center left columnspacing 0px end attributes row blank blank. end table 

Roboguru

Hasil dari  adalah ....

Pembahasan Soal:

Ingat kembali bahwa 

begin mathsize 14px style integral 1 over straight x space dx equals ln space vertical line straight x vertical line plus straight C end style   

Maka,

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral open parentheses fraction numerator 1 plus 2 straight x over denominator straight x end fraction close parentheses dx end cell equals cell integral open parentheses 1 over straight x plus fraction numerator 2 straight x over denominator straight x end fraction close parentheses dx end cell row blank equals cell integral 1 over straight x dx plus 2 space dx end cell row blank equals cell ln space vertical line straight x vertical line plus 2 straight x plus straight C end cell end table end style    

Jadi, jawaban yang benar adalah C.

 

Roboguru

Tentukan nilai dari: c.

Pembahasan Soal:

Rumus integral tak tentu:

integral a x to the power of n space d x equals fraction numerator a over denominator n plus 1 end fraction x to the power of n plus 1 end exponent plus c

integral 1 over x space d x equals ln space open vertical bar x close vertical bar plus c

integral e to the power of x space d x equals e to the power of x plus c

Integral pada soal di atas dapat ditentukan sebagai berikut.

integral open parentheses 5 e to the power of x plus 1 third x cubed minus 4 over x close parentheses space d x equals 5 e to the power of x plus 1 third times 1 fourth x to the power of 4 minus 4 space ln space open vertical bar x close vertical bar plus c equals 5 e to the power of x plus 1 over 12 x to the power of 4 minus 4 space ln space open vertical bar x close vertical bar plus c

Dengan demikian, integral open parentheses 5 e to the power of x plus 1 third x cubed minus 4 over x close parentheses space d x equals 5 e to the power of x plus 1 over 12 x to the power of 4 minus 4 space ln space open vertical bar x close vertical bar plus c 

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