Iklan

Pertanyaan

Hasil dari adalah ...

Hasil dari begin mathsize 12px style integral open parentheses 8 x minus 6 close parentheses square root of 2 x squared minus 3 x minus 5 end root space d x end style adalah ...

  1. begin mathsize 12px style 1 over 6 open parentheses 2 x squared minus 3 x minus 5 close parentheses square root of 2 x squared minus 3 x minus 5 end root plus c end style

  2. begin mathsize 12px style 1 third open parentheses 2 x squared minus 3 x minus 5 close parentheses square root of 2 x squared minus 3 x minus 5 end root plus c end style

  3. begin mathsize 12px style 2 over 3 open parentheses 2 x squared minus 3 x minus 5 close parentheses square root of 2 x squared minus 3 x minus 5 end root plus c end style

  4. begin mathsize 12px style 3 over 4 open parentheses 2 x squared minus 3 x minus 5 close parentheses square root of 2 x squared minus 3 x minus 5 end root plus c end style

  5. begin mathsize 12px style 4 over 3 open parentheses 2 x squared minus 3 x minus 5 close parentheses square root of 2 x squared minus 3 x minus 5 end root plus c end style

Ikuti Tryout SNBT & Menangkan E-Wallet 100rb

Habis dalam

01

:

23

:

23

:

02

Klaim

Iklan

R. RGFLSATU

Master Teacher

Jawaban terverifikasi

Pembahasan

Misalkan , maka , kemudian dengan menggunakan integral subsitusi akan didapat: Substitusikan kembali nilai t ke awal, sehingga diperoleh:

Misalkan begin mathsize 12px style t equals 2 x squared minus 3 x minus 5 end style, maka begin mathsize 12px style fraction numerator d t over denominator d x end fraction equals 4 x minus 3 rightwards arrow d x equals fraction numerator 1 over denominator 4 x minus 3 end fraction d t end style, kemudian dengan menggunakan integral subsitusi akan didapat:

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell integral open parentheses 8 x minus 6 close parentheses square root of 2 x squared minus 3 x minus 5 end root space d x end cell row blank equals cell integral 2 open parentheses 4 x minus 3 close parentheses square root of t fraction numerator 1 over denominator 4 x minus 3 end fraction d t end cell row blank equals cell integral 2 square root of t space d t end cell row blank equals cell 4 over 3 t square root of t plus c end cell end table

Substitusikan kembali nilai t ke awal, sehingga diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell integral open parentheses 8 x minus 6 close parentheses square root of 2 x squared minus 3 x minus 5 end root space d x end cell row blank equals cell 4 over 3 open parentheses 2 x squared minus 3 x minus 5 close parentheses square root of 2 x squared minus 3 x minus 5 end root plus c end cell end table

Perdalam pemahamanmu bersama Master Teacher
di sesi Live Teaching, GRATIS!

12

Iklan

Pertanyaan serupa

Hasil dari ∫ ( 2 x 2 + 6 x + 5 ) d x adalah...

1

0.0

Jawaban terverifikasi

RUANGGURU HQ

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

Coba GRATIS Aplikasi Roboguru

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

Hubungi Kami

Ruangguru WhatsApp

+62 815-7441-0000

Email info@ruangguru.com

[email protected]

Contact 02130930000

02130930000

Ikuti Kami

©2025 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia