Please integrate arctan(x)/x using any special functions required . You may use functions of a complex or imaginary argument,
To integrate the function (\frac{\arctan(x)}{x}), we can use integration by parts. Let's set:
[ u = \arctan(x) \quad \text{and} \quad dv = \frac{1}{x} , dx ]
Then, we differentiate and integrate to find (du) and (v):
[ du = \frac{1}{1+x^2} , dx \quad \text{and} \quad v = \ln|x| ]
Now, applying integration by parts:
[ \int \frac{\arctan(x)}{x} , dx = u v - \int v , du ]
Substituting in our expressions for (u), (v), (du):
[ = \arctan(x) \ln|x| - \int \ln|x| \cdot \frac{1}{1+x^2} , dx ]
Now we need to evaluate the integral (\int \ln|x| \cdot \frac{1}{1+x^2} , dx). This integral does not have a simple elementary form, but it can be expressed in terms of special functions.
To evaluate (\int \ln|x| \cdot \frac{1}{1+x^2} , dx), we can use the following result:
[ \int \frac{\ln(x)}{1+x^2} , dx = \frac{1}{2} \ln^2(x) + \frac{1}{2} \arctan(x) \ln(x) + C ]
Thus, we can write:
[ \int \frac{\arctan(x)}{x} , dx = \arctan(x) \ln|x| - \left( \frac{1}{2} \ln^2(x) + \frac{1}{2} \arctan(x) \ln(x) \right) + C ]
Combining the terms, we have:
[ \int \frac{\arctan(x)}{x} , dx = \arctan(x) \ln|x| - \frac{1}{2} \ln^2(x) - \frac{1}{2} \arctan(x) \ln(x) + C ]
This is the integral of (\frac{\arctan(x)}{x}) expressed in terms of logarithmic and arctangent functions.
Thus, the final result is:
[ \int \frac{\arctan(x)}{x} , dx = \arctan(x) \ln|x| - \frac{1}{2} \ln^2(x) - \frac{1}{2} \arctan(x) \ln(x) + C ]