Pertanyaan

∫ s + 3 s d s

$\int \frac{s d s}{s + 3}$

F. Kartikasari

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Jawaban

.

$integral fraction numerator s space d s over denominator square root of s plus 3 end root end fraction equals 2 over 3 open parentheses s plus 3 close parentheses to the power of 3 over 2 end exponent minus 6 open parentheses s plus 3 close parentheses to the power of 1 half end exponent plus c$ .

Pembahasan

Pertanyaan dari soal tersebut adalah menentukan hasil dari integral . Kita gunakan konsep integral substitusi. Rumus umum integral (anti turunan) sebagai berikut. Kita buat permisalan dari integral tersebut. Misal, Sehingga, Kemudian kita kembalikan ke permisalannya menjadi . Jadi, .

Pertanyaan dari soal tersebut adalah menentukan hasil dari integral $integral fraction numerator s space d s over denominator square root of s plus 3 end root end fraction$ . Kita gunakan konsep integral substitusi.

Rumus umum integral (anti turunan) sebagai berikut.

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

Kita buat permisalan dari integral tersebut. Misal,

table attributes columnalign right center left columnspacing 0px end attributes row x equals cell s plus 3 left right arrow x minus 3 equals s end cell row cell d x end cell equals cell d s end cell end table

Sehingga,

$table attributes columnalign right center left columnspacing 0px end attributes row cell integral fraction numerator s space d s over denominator square root of s plus 3 end root end fraction end cell equals cell integral fraction numerator x minus 3 space d x over denominator square root of x end fraction end cell row blank equals cell integral fraction numerator x over denominator square root of x end fraction d x minus integral fraction numerator 3 over denominator square root of x end fraction d x end cell row blank equals cell integral x to the power of 1 half end exponent d x minus integral 3 x to the power of negative 1 half end exponent d x end cell row blank equals cell fraction numerator x to the power of begin display style 1 half end style plus 1 end exponent over denominator begin display style 1 half end style plus 1 end fraction minus fraction numerator 3 x to the power of negative begin display style 1 half end style plus 1 end exponent over denominator negative begin display style 1 half end style plus 1 end fraction end cell row blank equals cell fraction numerator x to the power of begin display style 3 over 2 end style end exponent over denominator begin display style 3 over 2 end style end fraction minus fraction numerator 3 x to the power of begin display style 1 half end style end exponent over denominator begin display style 1 half end style end fraction end cell row blank equals cell 2 over 3 x to the power of 3 over 2 end exponent minus 6 x to the power of 1 half end exponent plus c end cell end table$

Kemudian kita kembalikan ke permisalannya menjadi 2 over 3 open parentheses s plus 3 close parentheses to the power of 3 over 2 end exponent minus 6 open parentheses s plus 3 close parentheses to the power of 1 half end exponent plus c .

Jadi, $integral fraction numerator s space d s over denominator square root of s plus 3 end root end fraction equals 2 over 3 open parentheses s plus 3 close parentheses to the power of 3 over 2 end exponent minus 6 open parentheses s plus 3 close parentheses to the power of 1 half end exponent plus c$ .