How do you find #int arctanx #?

Answer 1

We use the Integral of Inverse Functions theorem.

If #f^-1# denotes a continuous inverse function then
#\int f^{-1}(t)dt= t f^{-1}(t)-F( f^{-1}(t))+"c"#
We let #f^-1(x)=arctanx# and #f(x)=tanx#. Then #F(x)=inttanxdx=ln|secx|# and #F(f^-1(x))=ln|secarctanx|=lnsqrt(1+(tanarctanx)^2)=ln(sqrt(1+x^2))=1/2ln(1+x^2)#

So

#intarctanxdx=xarctanx-1/2ln(1+x^2)+"c"#
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Answer 2

The answer is #=xarctanx-1/2ln(1+x^2)+C#

Execute this integral through part-by-part integration.

#intuv'=uv-intu'v#
#u=arctanx#, #=>#, #u'=1/(1+x^2)#
#v'=1#, #=>#, #v=x#

Consequently,

The essential component is

#intarctanxdx=xarctanx-int(xdx)/(1+x^2)#
#int(xdx)/(1+x^2)=1/2int(2xdx)/(1+x^2)=1/2ln(1+x^2)#

Lastly,

#intarctanxdx=xarctanx-1/2ln(1+x^2)+C#
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Answer 3

# xarc tanx-1/2ln(x^2+1)+C, or, #

# xarc tanx-lnsqrt(x^2+1)+C#.

Suppose that, #I=intarc tanxdx=int(arc tanx)(1)dx#

Applying the Integration by Parts (IBP) Rule as follows:

IBP : #intuv'dx=uv-intu'vdx#.
We take, #u=arc tanx, and, v'=1#.
#:. u'=1/(x^2+1), and, v=intv'dx=int1dx=x#.
#:. I=xarc tanx-intx/(x^2+1)dx#,
#=xarc tanx-1/2int(2x)/(x^2+1)dx#,
#=xarctan x-1/2int{d/dx(x^2+1)}/(x^2+1)dx#.
Since, #int(f'(x))/f(x)dx=ln|f(x)#,
#:. I=xarc tanx-1/2ln(x^2+1)+C, or, #
# I=xarc tanx-lnsqrt(x^2+1)+C#.#.
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Answer 4

To find the integral of arctan(x), you can use integration by parts. Let u = arctan(x) and dv = dx. Then, du = 1/(1 + x^2) dx and v = x. Applying the integration by parts formula ∫u dv = uv - ∫v du, you get:

∫arctan(x) dx = x arctan(x) - ∫x/(1 + x^2) dx.

To integrate x/(1 + x^2), you can use the substitution method. Let u = 1 + x^2, then du = 2x dx. This simplifies the integral to ∫x/(1 + x^2) dx = ∫1/u du = ln|u| = ln(1 + x^2).

Therefore, the final result is:

∫arctan(x) dx = x arctan(x) - ln(1 + x^2) + C,

where C is the constant of integration.

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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.

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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