# How do you integrate #int ln(1+x^2)# by parts from #[0,1]#?

Now solving the resulting integral:

Substitute in (1):

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To integrate (\int_0^1 \ln(1+x^2) , dx) by parts, follow these steps:

- Let (u = \ln(1+x^2)) and (dv = dx).
- Calculate the differentials: (du = \frac{2x}{1+x^2} dx) and (v = x).
- Apply integration by parts formula: (\int u , dv = uv - \int v , du).
- Compute (uv) and (\int v , du).
- Evaluate the integral over the specified range [0,1].

Here are the steps in detail:

- Let (u = \ln(1+x^2)) and (dv = dx).
- Calculate the differentials: (du = \frac{2x}{1+x^2} dx) and (v = x).
- Apply integration by parts formula: [ \int u , dv = uv - \int v , du ]
- Compute: [ uv = x \ln(1+x^2) \Bigg|_0^1 - \int_0^1 x \cdot \frac{2x}{1+x^2} , dx ]
- Evaluate the integral: [ \int_0^1 \frac{2x^2}{1+x^2} , dx = \int_0^1 \frac{2x^2+2-2}{1+x^2} , dx = \int_0^1 \frac{2(x^2+1)-2}{1+x^2} , dx ] [ = 2\int_0^1 \frac{dx}{1+x^2} - 2\int_0^1 \frac{dx}{1+x^2} = 0 ]
- So, the integral becomes: [ x \ln(1+x^2) \Bigg|_0^1 - 0 = 1 \ln(1+1) - 0 = \ln(2) ]

Therefore, (\int_0^1 \ln(1+x^2) , dx = \ln(2)).

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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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