Roboguru

Carilah ∫x2−13​dx!

Pertanyaan

Carilah integral fraction numerator 3 over denominator x squared minus 1 end fraction d x!

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 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

L. Sibuea

Mahasiswa/Alumni Universitas Riau

Terakhir diupdate 06 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

Pertanyaan yang serupa

Hitung. 2. ∫4xx2−16​dx​

Pembahasan Soal:

Diperoleh intergalnya yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell integral fraction numerator d x over denominator 4 x square root of x squared minus 16 end root end fraction end cell equals cell 1 fourth integral fraction numerator 1 over denominator x square root of x squared minus 16 end root end fraction d x end cell row blank equals cell 1 fourth integral fraction numerator 1 over denominator x square root of x squared minus 4 squared end root end fraction d x end cell end table

Ingatlah bahwa:

integral fraction numerator 1 over denominator x square root of x squared minus a squared end root end fraction d x equals 1 over a times asec open parentheses fraction numerator open vertical bar x close vertical bar over denominator a end fraction close parentheses

Kemudian diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 fourth integral fraction numerator 1 over denominator x square root of x squared minus 4 squared end root end fraction d x end cell equals cell 1 fourth times 1 fourth times asec open parentheses fraction numerator open vertical bar straight x close vertical bar over denominator 4 end fraction close parentheses plus straight C end cell row blank equals cell 1 over 16 times asec open parentheses fraction numerator open vertical bar straight x close vertical bar over denominator 4 end fraction close parentheses plus straight C end cell row blank equals cell fraction numerator asec open parentheses begin display style fraction numerator open vertical bar straight x close vertical bar over denominator 4 end fraction end style close parentheses over denominator 16 end fraction plus straight C end cell end table

Maka, hasil integral dari integral fraction numerator d x over denominator 4 x square root of x squared minus 16 end root end fraction adalah fraction numerator asec open parentheses begin display style fraction numerator open vertical bar x close vertical bar over denominator 4 end fraction end style close parentheses over denominator 16 end fraction plus C.

0

Roboguru

Hasil dari ∫x1+2x​dx 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.

 

0

Roboguru

Hitung. 3. ∫2−5x2​dx​

Pembahasan Soal:

Diperoleh intergalnya yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell integral fraction numerator d x over denominator square root of 2 minus 5 x squared end root end fraction end cell equals cell integral fraction numerator 1 over denominator square root of 5 open parentheses begin display style 2 over 5 end style minus x squared close parentheses end root end fraction d x end cell row blank equals cell integral fraction numerator 1 over denominator square root of 5 times square root of begin display style 2 over 5 end style minus x squared end root end fraction d x end cell row blank equals cell fraction numerator 1 over denominator square root of 5 end fraction times integral fraction numerator 1 over denominator square root of begin display style 2 over 5 end style minus x squared end root end fraction d x end cell row blank equals cell fraction numerator 1 over denominator square root of 5 end fraction times fraction numerator square root of 5 over denominator square root of 5 end fraction times integral fraction numerator 1 over denominator square root of begin display style 2 over 5 end style minus x squared end root end fraction d x end cell row blank equals cell fraction numerator square root of 5 over denominator 5 end fraction times integral fraction numerator 1 over denominator square root of begin display style 2 over 5 end style minus x squared end root end fraction d x end cell end table

Ingatlah bahwa:

integral fraction numerator 1 over denominator a squared minus x squared end fraction d x equals arcsin times x over a

Kemudian diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator square root of 5 over denominator 5 end fraction integral fraction numerator 1 over denominator square root of begin display style 2 over 5 end style minus x squared end root end fraction d x end cell equals cell fraction numerator square root of 5 over denominator 5 end fraction times arcsin times fraction numerator x over denominator square root of begin display style 2 over 5 end style end root end fraction plus C end cell row blank equals cell fraction numerator square root of 5 over denominator 5 end fraction times arcsin times fraction numerator x over denominator begin display style fraction numerator square root of 2 over denominator square root of 5 end fraction end style end fraction plus C end cell row blank equals cell fraction numerator square root of 5 over denominator 5 end fraction times arcsin times x times fraction numerator square root of 5 over denominator square root of 2 end fraction plus C end cell row blank equals cell fraction numerator square root of 5 over denominator 5 end fraction times arcsin times fraction numerator x square root of 5 over denominator square root of 2 end fraction times fraction numerator square root of 2 over denominator square root of 2 end fraction plus C end cell row blank equals cell fraction numerator square root of 5 space arcsin space begin display style fraction numerator straight x square root of 10 over denominator 2 end fraction end style over denominator 5 end fraction plus C end cell end table

Maka, hasil integral dari integral fraction numerator d x over denominator square root of 2 minus 5 x squared end root end fraction adalah fraction numerator square root of 5 space arcsin space begin display style fraction numerator x square root of 10 over denominator 2 end fraction end style over denominator 5 end fraction plus C.

0

Roboguru

Hitung. 1. ∫9x2+16dx​

Pembahasan Soal:

Diperoleh intergalnya yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell integral fraction numerator d x over denominator 9 x squared plus 16 end fraction end cell equals cell integral fraction numerator d x over denominator 9 open parentheses x squared plus begin display style 16 over 9 end style close parentheses end fraction end cell row blank equals cell 1 over 9 integral fraction numerator 1 over denominator x squared plus begin display style 16 over 9 end style end fraction d x end cell end table

Ingatlah bahwa:

integral fraction numerator 1 over denominator x squared plus a squared end fraction d x equals 1 over a times arctan times x over a

Kemudian diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 over 9 integral fraction numerator 1 over denominator x squared plus begin display style 16 over 9 end style end fraction d x end cell equals cell 1 over 9 integral fraction numerator 1 over denominator x squared plus begin display style open parentheses 4 over 3 close parentheses squared end style end fraction d x end cell row blank equals cell 1 over 9 times fraction numerator 1 over denominator begin display style 4 over 3 end style end fraction times arctan times fraction numerator straight x over denominator begin display style 4 over 3 end style end fraction plus straight C end cell row blank equals cell 1 over 9 times 3 over 4 times arctan times straight x times 3 over 4 plus straight C end cell row blank equals cell 3 over 36 arctan fraction numerator 3 straight x over denominator 4 end fraction plus straight C end cell row blank equals cell fraction numerator arctan begin display style fraction numerator 3 straight x over denominator 4 end fraction end style over denominator 12 end fraction plus straight C end cell end table

Maka, hasil integral dari integral fraction numerator d x over denominator 9 x squared plus 16 end fraction adalah fraction numerator arctan begin display style fraction numerator 3 x over denominator 4 end fraction end style over denominator 12 end fraction plus C.

0

Roboguru

Tentukan nilai dari: b. ∫(2x1​+x2−ex)dx

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.

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell integral open parentheses fraction numerator 1 over denominator 2 x end fraction plus x squared minus e to the power of x close parentheses space d x end cell row blank equals cell integral open parentheses 1 half times 1 over x plus x squared minus e to the power of x close parentheses space d x end cell row blank equals cell 1 half space ln space open vertical bar x close vertical bar plus 1 third x cubed minus e to the power of x plus c end cell end table

Dengan demikian, integral open parentheses fraction numerator 1 over denominator 2 x end fraction plus x squared minus e to the power of x close parentheses space d x equals 1 half space ln space open vertical bar x close vertical bar plus 1 third x cubed minus e to the power of x plus c 

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved