فرمول های انتگرال (Integral) شامل عبارت $ {{\rm{a}}^2} - {x^2} $ ، در ریاضیات (Mathematics)
\[ \int {{1 \over {{a^2} - {x^2}}}dx} = {1 \over {2a}}\ln \left| {{{x + a} \over {x - a}}} \right| + C
\]
\[ \int {{1 \over {{{({a^2} - {x^2})}^2}}}dx} = {x \over {2{a^2}({a^2} - {x^2})}} + {1 \over {4{a^3}}}\ln \left| {{{x + a} \over {x - a}}} \right| + C
\]
\[ \int {{1 \over {\sqrt {{a^2} - {x^2}} }}dx} = {\sin ^{ - 1}}{x \over a} + C
\]
\[ \int {\sqrt {{a^2} - {x^2}} } dx = {x \over 2}\sqrt {{a^2} - {x^2}} + {{{a^2}} \over 2}{\sin ^{ - 1}}{x \over a} + C
\]
\[ \int {{x^2}} \sqrt {{a^2} - {x^2}} dx = {{{a^4}} \over 8}{\sin ^{ - 1}}{x \over a} - {1 \over 8}x\sqrt {{a^2} - {x^2}} ({a^2} - 2{x^2}) + C
\]
\[ \int {{{\sqrt {{a^2} - {x^2}} } \over x}} dx = \sqrt {{a^2} - {x^2}} - a\ln \left| {{{a + \sqrt {{a^2} - {x^2}} } \over x}} \right| + C
\]
\[ \int {{{\sqrt {{a^2} - {x^2}} } \over {{x^2}}}} dx = - {\sin ^{ - 1}}{x \over a} - {{\sqrt {{a^2} - {x^2}} } \over x} + C
\]
\[ \int {{{{x^2}} \over {\sqrt {{a^2} - {x^2}} }}} dx = {{{a^2}} \over 2}{\sin ^{ - 1}}{x \over a} - {1 \over 2}x\sqrt {{a^2} - {x^2}} + C
\]
\[ \int {{1 \over {x\sqrt {{a^2} - {x^2}} }}dx} = - {1 \over a}\ln \left| {{{a + \sqrt {{a^2} - {x^2}} } \over x}} \right| + C
\]
\[ \int {{1 \over {{x^2}\sqrt {{a^2} - {x^2}} }}dx} = - {{\sqrt {{a^2} - {x^2}} } \over {{a^2}x}} + C
\]
1
Thomas' Calculus Early Transcendentals - George B. Thomas Jr., Maurice D. Weir, Joel R. Hass - 13th Edition
نویسنده
علیرضا گلمکانی
شماره کلید
20018
گزینه ها
نظرات 0 0 0