∫√(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.
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?