# How do you find the indefinite integral of #int x^2/sqrt(x^3+3)dx#?

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To find the indefinite integral of ( \int \frac{x^2}{\sqrt{x^3 + 3}} , dx ), we can use the substitution method.

Let ( u = x^3 + 3 ). Then, ( du = 3x^2 , dx ).

Rewrite the integral in terms of ( u ): ( \frac{1}{3} \int \frac{1}{\sqrt{u}} , du ).

Now integrate ( \frac{1}{\sqrt{u}} ) with respect to ( u ): ( \int u^{-\frac{1}{2}} , du = 2u^{\frac{1}{2}} + C ).

Substitute back ( u = x^3 + 3 ): ( 2(x^3 + 3)^{\frac{1}{2}} + C ).

Therefore, the indefinite integral of ( \int \frac{x^2}{\sqrt{x^3 + 3}} , dx ) is ( 2(x^3 + 3)^{\frac{1}{2}} + C ), where ( C ) is the constant of integration.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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