# How do you find the indefinite integral of #int (x^4-6x^3+e^xsqrtx)/sqrtx dx#?

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To find the indefinite integral of (\int \frac{x^4 - 6x^3 + e^{x\sqrt{x}}}{\sqrt{x}} , dx), you can use the method of substitution. Let (u = \sqrt{x}), then (x = u^2) and (dx = 2u , du). Substituting these into the integral gives:

[ \int \frac{u^8 - 6u^6 + e^{u^3}}{u} \cdot 2u , du = \int (2u^7 - 12u^5 + 2u e^{u^3}) , du ]

Integrating term by term gives:

[ \int 2u^7 , du - \int 12u^5 , du + \int 2u e^{u^3} , du = \frac{2}{8}u^8 - \frac{12}{6}u^6 + \int 2u e^{u^3} , du ]

Simplifying, we get:

[ \frac{1}{4}u^8 - 2u^6 + \int 2u e^{u^3} , du ]

The remaining integral (\int 2u e^{u^3} , du) can be further simplified using another substitution. Let (v = u^3), then (du = \frac{dv}{3u^2}). Substituting into the integral gives:

[ \int 2u e^{u^3} , du = \int \frac{2e^v}{3} , dv = \frac{2}{3} \int e^v , dv = \frac{2}{3} e^v + C = \frac{2}{3} e^{u^3} + C ]

Substituting back in the original variable (x), the final result is:

[ \frac{1}{4}(\sqrt{x})^8 - 2(\sqrt{x})^6 + \frac{2}{3} e^{(\sqrt{x})^3} + C = \frac{1}{4}x^4 - 2x^3 + \frac{2}{3} e^{x\sqrt{x}} + 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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