Pertanyaan

Hasil dari ∫ x ( 2 x − 1 ) ( x + 2 ) d x adalah ...

Hasil dari $\int \frac{( 2 x - 1 ) ( x + 2 )}{x} d x$ adalah ... $\;$

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Habis dalam

Klaim

A. Armanda

Master Teacher

Mahasiswa/Alumni Universitas Indraprasta PGRI

Jawaban terverifikasi

Jawaban

.

$integral fraction numerator open parentheses 2 x minus 1 close parentheses open parentheses x plus 2 close parentheses over denominator square root of x end fraction space straight d x equals 4 over 5 x squared square root of x plus 2 x square root of x minus 4 square root of x plus C$ .

Pembahasan

Ingat! Maka, hasildari adalah: Jadi, .

Ingat!

$table attributes columnalign right center left columnspacing 0px end attributes row cell integral a x to the power of n space straight d x end cell equals cell fraction numerator a over denominator n plus 1 end fraction x to the power of n plus 1 end exponent plus C end cell row cell integral open parentheses f open parentheses x close parentheses plus-or-minus g open parentheses x close parentheses close parentheses space straight d x end cell equals cell integral f open parentheses x close parentheses space straight d x plus-or-minus integral g open parentheses x close parentheses space straight d x end cell row cell integral k times f open parentheses x close parentheses space straight d x end cell equals cell k integral f open parentheses x close parentheses space straight d x end cell end table$

Maka, hasil dari $integral fraction numerator open parentheses 2 x minus 1 close parentheses open parentheses x plus 2 close parentheses over denominator square root of x end fraction space straight d x$ adalah:

$table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell integral fraction numerator open parentheses 2 x minus 1 close parentheses open parentheses x plus 2 close parentheses over denominator square root of x end fraction space straight d x end cell row blank equals cell integral x to the power of negative 1 half end exponent open parentheses 2 x minus 1 close parentheses open parentheses x plus 2 close parentheses space straight d x end cell row blank equals cell integral open parentheses 2 x to the power of 1 half end exponent minus x to the power of negative 1 half end exponent close parentheses open parentheses x plus 2 close parentheses space straight d x end cell row blank equals cell integral 2 x to the power of 3 over 2 end exponent plus 3 x to the power of 1 half end exponent minus 2 x to the power of negative 1 half end exponent space d x end cell row blank equals cell 2 integral x to the power of 3 over 2 end exponent space d x plus 3 integral x to the power of 1 half end exponent space d x minus 2 integral x to the power of negative 1 half end exponent space d x end cell row blank equals cell fraction numerator 2 over denominator begin display style bevelled 5 over 2 end style end fraction x to the power of 5 over 2 end exponent plus fraction numerator 3 over denominator begin display style bevelled 3 over 2 end style end fraction x to the power of 3 over 2 end exponent minus fraction numerator 2 over denominator begin display style bevelled 1 half end style end fraction x to the power of 1 half end exponent plus C end cell row blank equals cell 4 over 5 x squared square root of x plus 2 x square root of x minus 4 square root of x plus C end cell end table$

Jadi, $integral fraction numerator open parentheses 2 x minus 1 close parentheses open parentheses x plus 2 close parentheses over denominator square root of x end fraction space straight d x equals 4 over 5 x squared square root of x plus 2 x square root of x minus 4 square root of x plus C$ .