# How do you find the integral of #ln(1+x^2)#?

The answer is

Utilizing integration by parts, determine this integral.

Consequently, the integral is

The subsequent integral is

Ultimately, the integral is

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To find the integral of ln(1+x^2), you can use the substitution method. Let ( u = 1 + x^2 ). Then, ( du/dx = 2x ), and ( dx = du/(2x) ). Substitute these into the integral to get:

[ \int \ln(1+x^2) , dx = \int \ln(u) \frac{du}{2x} ]

Using integration by parts, let ( dv = du/(2x) ) and ( u = \ln(u) ). Then, ( v = u \ln(u) - \int u , du/u ).

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

Now, use a trigonometric substitution, letting ( x = \tan(\theta) ). Then, ( dx = \sec^2(\theta) , d\theta ).

[ \int \ln(1+x^2) , dx = x \ln(1+x^2) - \frac{1}{2} \int \frac{2\tan^2(\theta) \sec^2(\theta) , d\theta}{1+\tan^2(\theta)} ]

[ = x \ln(1+x^2) - \frac{1}{2} \int \sin^2(\theta) , d\theta ]

[ = x \ln(1+x^2) - \frac{1}{4} (\theta - \sin(\theta) \cos(\theta)) + C ]

[ = x \ln(1+x^2) - \frac{1}{4} (\arctan(x) - x/(1+x^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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