$$\int_{-2}^{6} \frac{1}{\sqrt{1+(-x)^2}} \, dx$$
When performing this integral on paper, I get
$$\sinh^{-1}(6) - \sinh^{-1}(-2) $$
But when I type it on wolframalpha, I get the unintuitive answer
$$\sinh^{-1}(6) + \sinh^{-1}(2) \approx 3.935$$
http://www.wolframalpha.com/input/?i=integral%281%2Fsqrt%281%2B%28-x%29%5E2%29%29+from+-2+to+6
What's happening to the negative signs?
$\sinh^{-2} x$ is an odd function: $$\sinh^{-1}(x):=\ln(x+\sqrt{x^2+1})\\ \implies \sinh^{-1}(-x)=\ln(-x+\sqrt{x^2+1})=\ln(-(x-\sqrt{x^2+1}))=\ln\left(\dfrac{x-\sqrt{1+x^2}}{x^2-1-x^2}\right)=\\ \ln\left(\dfrac{1}{x-\sqrt{1+x^2}}\dfrac{x+\sqrt{1+x^2}}{x+\sqrt{1+x^2}}\right)=\ln\left(\dfrac{1}{x-\sqrt{1+x^2}}\right)=-\ln(x+\sqrt{x^2+1})\\ \implies \sinh^{-1}(-x)=-\sinh^{-1}(x)\\ \text{Thus, } -\sinh^{-1}(-2)=-(-\sinh^{-1}(2))=\sinh^{-1}(2)$$