Haqiqatan ham integralni baholashdan boshqa ko'p narsa qilish kerak emas.

\[\int{{\frac{1}{{1 + {x^2}}} + \frac{{12}}{{\sqrt {1 - {x^2}} }}\,dx}} = \require{bbox} \bbox[2pt,border:1px solid black]{{{{\tan }^{ - 1}}\left( x \right) + 12{{\sin }^{ - 1}}\left( x \right) + c}}\]

E'tibor bering, teskari sinus va teskari kosinus hosilasining o'xshashligi tufayli muqobil javob:

\[\int{{\frac{1}{{1 + {x^2}}} + \frac{{12}}{{\sqrt {1 - {x^2}} }}\,dx}} = \require{bbox} \bbox[2pt,border:1px solid black]{{{{\tan }^{ - 1}}\left( x \right) - 12{{\cos }^{ - 1}}\left( x \right) + c}}\]