acim
1) int x³/√(x² + 1) dx = int x(x²)/√(x²+1) dx misal u = x² + 1 -------> x² = u - 1 du = 2x dx 1/2 du = x dx maka integral di atas menjadi : = int 1/2 (u - 1)/√u du = 1/2 int u^(1/2) - u^(-1/2) du = 1/2 (2/3 u^(3/2) - 2u^(1/2)) + C = 1/3 u^(3/2) - u^(1/2) + C = 1/3 u√u - √u + C = 1/3 (x² + 1)√(x²+1) - √(x²+1) + C
2) int 2sin(x)cos(x) dx = int sin(2x) dx = -1/2 cos(2x) + C
3) int (x+2)³ dx = int (x³ + 6x² + 12x + 8) dx = 1/4 x^4 + 2x³ + 6x² + 8x + C
1 votes Thanks 1
wahyu032014
yang No2 dan No 1 kurang ngerti bisa dijelaskan secara rinci
acim
no . 1 menggunakan integral u-subs, sdh pernah belajar blom materi ini ?
no. 2 hanya menggunakan identitas trigono pada double angle formula
wahyu032014
no 2 alhamdulilah udah ngerti .. Integral subtitusi udah beljar tapi blum paham paham soal yg diakasih gurunya .. yg blum paham dari int 1/2 (U-1)/√u du
= int x(x²)/√(x²+1) dx
misal u = x² + 1 -------> x² = u - 1
du = 2x dx
1/2 du = x dx
maka integral di atas menjadi :
= int 1/2 (u - 1)/√u du
= 1/2 int u^(1/2) - u^(-1/2) du
= 1/2 (2/3 u^(3/2) - 2u^(1/2)) + C
= 1/3 u^(3/2) - u^(1/2) + C
= 1/3 u√u - √u + C
= 1/3 (x² + 1)√(x²+1) - √(x²+1) + C
2) int 2sin(x)cos(x) dx
= int sin(2x) dx
= -1/2 cos(2x) + C
3) int (x+2)³ dx
= int (x³ + 6x² + 12x + 8) dx
= 1/4 x^4 + 2x³ + 6x² + 8x + C