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Evaluate the integral by making the given substitution. (Use C for the constant of integration.)

∫ x^2√x^3+37 dx, u=x^3+37

Sagot :

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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

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