فرمول های انتگرال (Integral) شامل عبارت $ {x^2} - {{\rm{a}}^2} $ ، در ریاضیات (Mathematics)
\[ \int {{1 \over {\sqrt {{x^2} - {a^2}} }}dx} = \ln |x + \sqrt {{x^2} - {a^2}} | + C
\]
\[ \int {\sqrt {{x^2} - {a^2}} } dx = {x \over 2}\sqrt {{x^2} - {a^2}} - {{{a^2}} \over 2}\ln |x + \sqrt {{x^2} - {a^2}} | + C
\]
\[ \int {{{(\sqrt {{x^2} - {a^2}} )}^n}} dx = {{x{{(\sqrt {{x^2} - {a^2}} )}^n}} \over {n + 1}} - {{n{a^2}} \over {n + 1}}\int {{{(\sqrt {{x^2} - {a^2}} )}^{n - 2}}} dx \ \ \ \ ,n \ne - 1
\]
\[ \int {{1 \over {{{(\sqrt {{x^2} - {a^2}} )}^n}}}dx} = {{x{{(\sqrt {{x^2} - {a^2}} )}^{2 - n}}} \over {(2 - n){a^2}}} - {{n - 3} \over {(n - 2){a^2}}}\int {{{dx} \over {{{(\sqrt {{x^2} - {a^2}} )}^{n - 2}}}}} \ \ \ \ ,n \ne 2
\]
\[ \int x {(\sqrt {{x^2} - {a^2}} )^n}dx = {{{{(\sqrt {{x^2} - {a^2}} )}^{n + 2}}} \over {n + 2}} + C \ \ \ \ ,n \ne - 2
\]
\[ \int {{x^2}} \sqrt {{x^2} - {a^2}} dx = {x \over 8}(2{x^2} - {a^2})\sqrt {{x^2} - {a^2}} - {{{a^4}} \over 8}\ln |x + \sqrt {{x^2} - {a^2}} | + C
\]
\[ \int {{{\sqrt {{x^2} - {a^2}} } \over x}} dx = \sqrt {{x^2} - {a^2}} - a{\sec ^{ - 1}}\left| {{x \over a}} \right| + C
\]
\[ \int {{{\sqrt {{x^2} - {a^2}} } \over {{x^2}}}} dx = \ln |x + \sqrt {{x^2} - {a^2}} | - {{\sqrt {{x^2} - {a^2}} } \over x} + C
\]
\[ \int {{{{x^2}} \over {\sqrt {{x^2} - {a^2}} }}} dx = {{{a^2}} \over 2}\ln |x + \sqrt {{x^2} - {a^2}} | + {x \over 2}\sqrt {{x^2} - {a^2}} + C
\]
\[ \int {{1 \over {x\sqrt {{x^2} - {a^2}} }}dx} = {1 \over a}{\sec ^{ - 1}}\left| {{x \over a}} \right| + C = {1 \over a}{\cos ^{ - 1}}\left| {{a \over x}} \right| + C
\]
\[ \int {{1 \over {{x^2}\sqrt {{x^2} - {a^2}} }}dx} = {{\sqrt {{x^2} - {a^2}} } \over {{a^2}x}} + C
\]
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Thomas' Calculus Early Transcendentals - George B. Thomas Jr., Maurice D. Weir, Joel R. Hass - 13th Edition
نویسنده علیرضا گلمکانی
شماره کلید 20020
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