# How do you integrate #xln(1+x) dx#?

I would separate the integral and begin with a u-substitution.

Thus:

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To integrate (x \ln(1+x) , dx), you can use integration by parts. Let (u = \ln(1+x)) and (dv = x , dx). Then, (du = \frac{1}{1+x} , dx) and (v = \frac{1}{2}x^2). Applying the integration by parts formula (\int u , dv = uv - \int v , du), we have:

[ \begin{align*} \int x \ln(1+x) , dx &= \frac{1}{2}x^2 \ln(1+x) - \int \frac{1}{2}x^2 \cdot \frac{1}{1+x} , dx \ &= \frac{1}{2}x^2 \ln(1+x) - \frac{1}{2} \int \frac{x^2}{1+x} , dx \end{align*} ]

Now, to integrate (\int \frac{x^2}{1+x} , dx), use polynomial division or rewrite it as (\int \left(x - \frac{x}{1+x}\right) , dx). Then, integrate term by term:

[ \int \frac{x^2}{1+x} , dx = \int \left(x - \frac{x}{1+x}\right) , dx ]

[ = \frac{1}{2}x^2 - \int \frac{x}{1+x} , dx ]

For the integral (\int \frac{x}{1+x} , dx), perform a simple substitution or use (u)-substitution with (u = 1+x). This results in:

[ \int \frac{x}{1+x} , dx = \frac{1}{2}x^2 - \ln|1+x| + C ]

Substitute this back into the original equation:

[ \int x \ln(1+x) , dx = \frac{1}{2}x^2 \ln(1+x) - \frac{1}{2}\left(\frac{1}{2}x^2 - \ln|1+x|\right) + C ]

Simplify this expression to obtain the final answer.

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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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