# How do you integrate #int_-1^1x(1+x)^3dx#?

Do a "u" substitute

Change the limits:

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To integrate (\int_{-1}^{1}x(1+x)^3 , dx), you can use the method of integration by parts. Let (u = x) and (dv = (1+x)^3 , dx). Then, (du = dx) and (v = \frac{1}{4}(1+x)^4). Applying the integration by parts formula (\int u , dv = uv - \int v , du), we get:

[ \begin{aligned} \int x(1+x)^3 , dx &= \frac{1}{4}x(1+x)^4 - \int \frac{1}{4}(1+x)^4 , dx \ &= \frac{1}{4}x(1+x)^4 - \frac{1}{4} \cdot \frac{1}{5}(1+x)^5 + C, \end{aligned} ]

where (C) is the constant of integration. Finally, evaluate the definite integral from (-1) to (1) by substituting the limits:

[ \begin{aligned} \int_{-1}^{1}x(1+x)^3 , dx &= \left[\frac{1}{4}x(1+x)^4 - \frac{1}{4} \cdot \frac{1}{5}(1+x)^5\right]_{-1}^{1} \ &= \left(\frac{1}{4}(1)(2)^4 - \frac{1}{4} \cdot \frac{1}{5}(2)^5\right) - \left(\frac{1}{4}(-1)(0)^4 - \frac{1}{4} \cdot \frac{1}{5}(0)^5\right) \ &= \frac{1}{4}(2^5) - \frac{1}{4} \cdot \frac{1}{5}(2^5) \ &= \frac{31}{5}. \end{aligned} ]

So, the value of the integral (\int_{-1}^{1}x(1+x)^3 , dx) is (\frac{31}{5}).

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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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