Integrate [rad arctan (2x)]/ (1+ 4x^2) dx
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Integrate [rad arctan (2x)]/ (1+ 4x^2) dx

[From: ] [author: ] [Date: 11-12-16] [Hit: ]
-rad arctan (2x) Whatsrad? You sure?......
∫√(arctan (2x))/ (1+ 4x^2) dx

Lest see something, lets derivate arctan(2x)

y=arctan(2x)


y'= 2/(1+(2x)²)

y'=2/(1+4x²)

Soo as you can see, the derivative of arctan(2x) is present in the integral, so lets make a simple subtitution.

∫√(arctan (2x))/ (1+ 4x^2) dx

u=arctan(2x)

du=2/(1+4x²)*dx

du/2=1/(1+4x²)*dx

Replacing.

1/2∫√u du

Integrating.

Having in mind that √u =u½

1/2(2/3*(u)^3/2)+K

1/3*(u)^(3/2) +K

Substitution back u=arctan(2x)

1/3*(arctan(2x))^(3/2) +K

That is your answer.



Remember ^(3/2) Is raised to the 3/2

Hope you understand.

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"rad arctan (2x)" What's "rad"? You sure?
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keywords: arctan,rad,Integrate,dx,Integrate [rad arctan (2x)]/ (1+ 4x^2) dx
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