# Evaluation of $\int_{1}^{\infty} \frac{\arctan (x+1)}{x^2+4} dx$ in closed form? [duplicate]

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 ?

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.