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How do you find the antiderivative of #int (xarctanx) dx#?

How do you find the antiderivative : #int (xarctanx) dx#?

Answer 1

Christopher Agresta

# 1/2{(x^2+1)arc tanx-x}+C.#

Suppose that, #I=intx arc tanxdx.#

We will use the following Method of Integration by Parts (IBP) :

IBP :#intu*vdx=uintvdx-int((du)/dxintvdx)dx.#

We take, #u=arc tanx, and, v=x.#

#:. (du)/dx=1/(x^2+1), and, intvdx=x^2/2.#

Therefore, by IBP, we have,

#I=x^2/2*arc tanx-int{x^2/2*1/(x^2+1)}dx,#

#=x^2/2*arc tanx-1/2intx^2/(x^2+1)dx,#

#=x^2/2*arc tanx-1/2int{(x^2+1)-1}/(x^2+1)dx,#

#=x^2/2*arc tanx-1/2int{(x^2+1)/(x^2+1)-1/(x^2+1)}dx,#

#=x^2/2*arc tanx-1/2{int1dx-int1/(x^2+1)dx},#

#=x^2/2*arc tanx-1/2{x-arc tanx}.#

# rArr I=1/2{(x^2+1)arc tanx-x}+C.#

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Answer 2

Luke Alvarez

To find the antiderivative of ∫(xarctanx) dx, you can use integration by parts. Let u = arctan(x) and dv = x dx. Then, differentiate u to find du and integrate dv to find v. After that, apply the integration by parts formula:

∫udv = uv - ∫vdu

Finally, substitute the values of u, v, du, and dv into the formula and solve for the antiderivative.

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Answer from HIX Tutor

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.