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Hasil dari ∫ ( 6 x 2 + 4 x ) x 3 + x 2 − 7 ​ d x = ...

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  1. 2 over 3 cube root of open parentheses x cubed plus x squared minus 7 close parentheses squared end root plus C

  2. 2 over 3 square root of open parentheses x cubed plus x squared minus 7 close parentheses cubed end root plus C

  3. 4 over 3 square root of open parentheses x cubed plus x squared minus 7 close parentheses cubed end root plus C

  4. 4 over 3 cube root of open parentheses x cubed plus x squared minus 7 close parentheses squared end root plus C

  5. 4 over 3 square root of open parentheses x cubed plus x squared minus 7 close parentheses end root plus C

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K. Putri

Master Teacher

Mahasiswa/Alumni Universitas Pendidikan Ganesha

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Pembahasan

integral open parentheses 6 x squared plus 4 x close parentheses square root of x cubed plus x squared minus 7 end root space d x  equals 2 integral open parentheses 3 x squared plus 2 x close parentheses open parentheses x cubed plus x squared minus 7 close parentheses to the power of 1 half end exponent space d x  equals 2 integral open parentheses x cubed plus x squared minus 7 close parentheses to the power of 1 half end exponent space d open parentheses x cubed plus x squared minus 7 close parentheses  equals 2.1 third open parentheses x cubed plus x squared minus 7 close parentheses to the power of 3 over 2 end exponent plus C  equals 4 over 3 square root of open parentheses x cubed plus x squared minus 7 close parentheses cubed end root space plus C

Latihan Bab

Pengenalan Integral

Integral Tak Tentu

Integral Substitusi

Aplikasi Integral Tak Tentu

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