Daftar Isi
hasil dr integral (x+1)(x2+2x-3)4 dx=
Jawaban:
Rumus integral –
[tex] \tt \int ax ^ m dx = \frac a m + 1 x ^ m + 1 + C \\ [/tex]
Rumus integral (cepat untuk perkalian pangkat tinggi)
[tex] \tt \int f(x).h(x)^ n dx = \frac f(x) . h(x)^ n + 1 (n + 1) . h'(x) + C \\ [/tex]
[tex] \\ [/tex]
Ada dua cara dalam penyelesaian ini
Pertama, saya pakai rumus yg diatas
[tex] \sf \int(x + 1)(x ^ 2 + 2x – 3) ^ 4 dx[/tex]
[tex]= \sf \frac (x + 1)(x ^ 2 + 2x – 3) ^ 4 + 1 (4 + 1).(2x + 2) + C[/tex]
[tex]= \sf \frac (x + 1)(x ^ 2 + 2x – 3) ^ 5 5(2x + 2) + C[/tex]
[tex] = \sf\frac (x ^ 2 + 2x – 3) ^ 5 5.2 + C[/tex]
[tex]= \sf \frac (x ^ 2 + 2x – 3) 10 + C[/tex]
[tex] \\ [/tex]
kedua, saya pakai integral subtitusi
misal:
- k = (x² + 2x – 3)
- dk/dx = 2x + 2
- dx = 1/(2x + 2) dk
[tex] \sf \int(x + 1) \: . \: k ^ 4 \: . \: \frac 1 2x + 2 dk \: [/tex]
[tex] \sf \int k ^ 4 \: . \: \frac x + 1 2x + 2 dk[/tex]
[tex] = \sf \frac k ^ 4 + 1 4 + 1 . \frac 1 2 + C [/tex]
[tex] = \sf \frac k ^ 5 5 . \frac 1 2 + C [/tex]
[tex] = \sf\frac k ^ 5 10 + C [/tex]
ganti k dgn (x² + 2x – 3)
[tex] = \sf \frac (x ^ 2 + 2×3) ^ 5 10 + C [/tex]
[tex] \\ [/tex]
Kesimpulan
- Jadi, hasil dr [tex] \sf \int(x + 1)(x ^ 2 + 2x – 3) ^ 4 dx [/tex] adalah [tex] \sf \frac (x ^ 2 + 2x – 3) 10 + C [/tex]
[tex] \\ [/tex]
Detail jawaban
♬ Mapel : Matematika
♬ Kelas : XI
♬ Materi : Bab 10 – Integral tak tentu
♬ Kode mapel : 2
♬ Kode kategorisasi : 11.2.10
♬ Kata kunci : Integral
____________________________
Semangattt ya’
hasil dr integral (x+1)(x2+2x-3)3 dx=
Jawaban:
1/8 (x² + 2x -3)^4 + C
Penjelasan dgn langkah-langkah:
caranya lihat foto atas
Hitung integral tak tentu ∫(x – 1)/(x2 – 2x – 3)dx, kemudian hitung pula 4∫5(x – 1)/(x2 – 2x – 3)dx
Jawaban:
itu saya kasih contohnya ya
integral 8x(x2 +5)3 dx =
integral 14x per (x2-3)8 =
integral 2x akar x2+4 dx =
integral 12×2(x3-4)5 dx =
integral (2x+3)(x2+3x-1)4 dx =
integral 6x+1 per (3×2+x-2)4 dx =
integral 8x per (x2+1) dx =
integral 4x.e^x2 dx =
Penjelasan dgn langkah-langkah:
1)
[tex]integral \: 8x( x ^ 2 + 5) ^ 3 dx[/tex]
misal
[tex]u = x ^ 2 + 5 \\ du = 2x \: dx \\ dx = \frac du 2x [/tex]
[tex] = integral \: 8x \: u ^ 3 \frac du 2x \\ = 4 \:( \frac u ^ (3 + 1) 3 + 1 ) + c\\ = u ^ 4 + c \\ = ( x ^ 2 + 5) ^ 4 + c[/tex]
2)
[tex]integral \: \frac 14x ( x ^ 2 – 3 )^ 8 dx[/tex]
misal
[tex]u = x ^ 2 – 3 \\ du = 2x \: dx \\ dx = \frac du 2x [/tex]
[tex] = integral \: \frac 14x u^ 8 \frac du 2x [/tex]
[tex] =7 \: integral \: u^ 8 du[/tex]
[tex] = 7( \frac u ^ (8 + 1) 8 + 1 ) + c \\ = \frac 7 9 u ^ 9 + c \\ = \frac 7 9 ( x ^ 2 – 3) ^ 9 + c[/tex]
3)
[tex] = integral \:2x \sqrt x ^ 2 + 4 \: dx[/tex]
misal
[tex]u = x ^ 2 + 4 \\ du = 2x \: dx \\ dx = \frac du 2x [/tex]
[tex] = integral \:2x(u) ^ \frac 1 2 \: \frac du 2x [/tex]
[tex] = integral \:(u) ^ \frac 1 2 \: du[/tex]
[tex] = \frac u ^ ( \frac 1 2 + \frac 2 2 ) \frac 1 2 + \frac 2 2 + c \\ = \frac u ^ \frac 3 2 \frac 3 2 + c \\ = \frac 2 3 ( x ^ 2 + 4) \sqrt x ^ 2 + 4 + c[/tex]
4)
[tex]integral \: 12 x ^ 2 ( x ^ 3 – 4) ^ 5 dx[/tex]
misal
[tex]u = x ^ 3 – 4 \\ du = 3 x ^ 2 dx \\ dx = \frac du 3 x ^ 2 [/tex]
[tex] = integral \: 12 x ^ 2 ( u) ^ 5 \frac du 3 x ^ 2 [/tex]
[tex] =4 \: integral \: ( u) ^ 5 du[/tex]
[tex] = 4( \frac u ^ 6 6 ) + c \\ = \frac 4 6 ( x ^ 3 – 4) ^ 6 + c[/tex]
[tex] = \frac 2 3 ( x ^ 3 – 4) ^ 6 + c[/tex]
5)
[tex]integral \: (2x + 3)( x ^ 2 + 3x – 1) ^ 4 dx[/tex]
misal
[tex]u = x ^ 2 + 3x – 1 \\ du =( 2x + 3)dx \\ dx = \frac du 2x + 3 [/tex]
[tex] = integral \: (2x + 3)( u) ^ 4 \frac du 2x + 3 [/tex]
[tex] = integral \: ( u) ^ 4 du[/tex]
[tex] = \frac u ^ 5 5 + c \\ = \frac 1 5 ( x ^ 2 + 3x – 1) ^ 5 + c[/tex]
6)
[tex] integral \: \frac 6x + 1 (3 x ^ 2 + x – 2) ^ 4 dx[/tex]
misal
[tex]u = 3 x ^ 2 + x – 2 \\ du =( 6x + 1)dx \\ dx = \frac du 6x + 1 [/tex]
[tex] = integral \: \frac 6x + 1 (u) ^ 4 \frac du 6x + 1 [/tex]
[tex] = integral \: (u) ^ – 4 du[/tex]
[tex] = \frac u ^ – 3 – 3 + c \\ = – \frac 1 3 u ^ 3 + c \\ = – \frac 1 3(3 x ^ 2 + x – 1) ^ 3 + c[/tex]
7)
[tex]integral \: \frac 8x x ^ 2 + 1 dx[/tex]
misal
[tex]u = x ^ 2 + 1 \\ du = 2x \: dx \\ dx = \frac du 2x [/tex]
[tex] = integral \: \frac 8x u \frac du 2x [/tex]
[tex] =4 \: integral \: u ^ – 1 du[/tex]
[tex] = 4 \: ln \: u + c \\ = 4 \: ln \: ( x ^ 2 + 1) + c[/tex]
a. (2x-1) (x2-x+3)3 dx
tentukan hasil dr integral
penyelesaian terlampir ya. 