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I have used the substitution $u= x+1 $ then $du =dx$ to evaluate the following integral in closed form and since $\arctan$ is connected to $x^2+1$ as it is a derivative of it. $$\int_{1}^{\infty} \frac{\arctan (x+1)}{x^2+4} dx$$ But i didn't come up to it's closed form the inverse symbolic calculator doesn't give me the closed form for it , and all my attempts gives me this approach : $$\int_{1}^{\infty} \frac{\arctan (x+1)}{x^2+4} dx=\frac{\pi^2}{8}-\frac{3\pi}{8}\arctan \left(\frac{1}{2}\right)$$ But this is not the same result with Wolfram alpha, Any way ?

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I suppose a mistake somewhere since $$\frac{\pi^2}{8}-\frac{3\pi}{8}\arctan \big(\frac{1}{2})=0.687479$$ while Wolfram Alpha returns a value equal to $0.741221$ which is correct.

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