# How do solve #∫x tan^-1x dx#, given that #d/dx tan^-1x = 1/(1+x^2)# ?

Then:

Simplifying slightly:

Which we may write as:

Rearranging and adding the constant of integration:

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To solve ∫x tan^-1x dx, use integration by parts, where u = tan^-1x and dv = x dx.

This gives: du = (1/(1+x^2)) dx v = (1/2)x^2

Apply integration by parts formula:

∫u dv = uv - ∫v du

Substitute the values of u, v, du, and dv:

= (x * tan^-1x * (1/2)x^2) - ∫((1/2)x^2 * (1/(1+x^2))) dx

= (x^3/2) * tan^-1x - (1/2) * ∫(x^2/(1+x^2)) dx

Now, let's solve the remaining integral by using a substitution method.

Let u = x^2 + 1, du = 2x dx Thus, dx = (1/2) du

Substituting, we get:

= (x^3/2) * tan^-1x - (1/4) * ∫(1/u) du

= (x^3/2) * tan^-1x - (1/4) * ln|u| + C

Finally, substitute back x^2 + 1 for u:

= (x^3/2) * tan^-1x - (1/4) * ln|x^2 + 1| + C

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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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