The Given integral is :
∫x²√x³ + 37 dx ; u = x³ + 37
Answer:
2/9(x³ + 37)^3/2 + C
Step-by-step explanation:
∫x²√x³ + 37 dx ; u = x³ + 37
∫x²√u dx
Since u = x³ + 37
Differentiate u with respect to x
du/dx = 3x²
du = 3x²dx
dx = du / 3x²
∫x²√u dx = ∫x²√u du/3x²
1/3x²∫x²√u du
1/3 ∫ √u du
Integrating √u with respect to u
1/3 ∫ u^1/2 du
1/3 [u^(1/2+1) / 1/2 + 1] + C
1/3 [u^3/2 / 3/2] + C
1/3 u^3/2 * 2/3 + C
1/3*2/3 u^3/2 + C
2/9 u^3/2
u = x³ + 37
2/9(x³ + 37)^3/2 + C
