فرمول های انتگرال (Integral) شامل عبارت $ \mathrm{a}x+b $ ، در ریاضیات (Mathematics)
\[ \int {{{(ax + b)}^n}} dx = {{{{(ax + b)}^{n + 1}}} \over {a(n + 1)}} + C \ \ \ \ ,n \ne - 1
\]
\[ \int x {(ax + b)^n}dx = {{{{(ax + b)}^{n + 1}}} \over {{a^2}}}\left[ {{{ax + b} \over {n + 2}} - {b \over {n + 1}}} \right] + C \ \ \ \ ,n \ne - 1, - 2
\]
\[ \int {{{(ax + b)}^{ - 1}}} dx = {1 \over a}\ln |ax + b| + C
\]
\[ \int x {(ax + b)^{ - 1}}dx = {x \over a} - {b \over {{a^2}}}\ln |ax + b| + C
\]
\[ \int x {(ax + b)^{ - 2}}dx = {1 \over {{a^2}}}\left[ {\ln |ax + b| + {b \over {ax + b}}} \right] + C
\]
\[ \int {{{dx} \over {x(ax + b)}}} = {1 \over b}\ln |{x \over {ax + b}}| + C
\]
\[ \int {{{(\sqrt {ax + b} )}^n}} dx = {2 \over a}{{{{(\sqrt {ax + b} )}^{n + 2}}} \over {n + 2}} + C \ \ \ \ ,n \ne - 2
\]
\[ \int {{{\sqrt {ax + b} } \over x}} dx = 2\sqrt {ax + b} + b\int {{{dx} \over {x\sqrt {ax + b} }}}
\]
\[ \int {{{dx} \over {x\sqrt {ax + b} }}} = {1 \over {\sqrt b }}\ln \left| {{{\sqrt {ax + b} - \sqrt b } \over {\sqrt {ax + b} + \sqrt b }}} \right| + C
\]
\[ \int {{{dx} \over {x\sqrt {ax - b} }}} = {2 \over {\sqrt b }}{\tan ^{ - 1}}\sqrt {{{ax - b} \over b}} + C
\]
\[ \int {{{\sqrt {ax + b} } \over {{x^2}}}} dx = - {{\sqrt {ax + b} } \over x} + {a \over 2}\int {{{dx} \over {x\sqrt {ax + b} }}} + C
\]
\[ \int {{{dx} \over {{x^2}\sqrt {ax + b} }}} = - {{\sqrt {ax + b} } \over {bx}} - {a \over {2b}}\int {{{dx} \over {x\sqrt {ax + b} }}} + C
\] \[ \int {{{(ax + b)}^n}} dx = {{{{(ax + b)}^{n + 1}}} \over {a(n + 1)}} + C \ \ \ \ ,n \ne - 1
\]
\[ \int x {(ax + b)^n}dx = {{{{(ax + b)}^{n + 1}}} \over {{a^2}}}\left[ {{{ax + b} \over {n + 2}} - {b \over {n + 1}}} \right] + C \ \ \ \ ,n \ne - 1, - 2
\]
\[ \int {{{(ax + b)}^{ - 1}}} dx = {1 \over a}\ln |ax + b| + C
\]
\[ \int x {(ax + b)^{ - 1}}dx = {x \over a} - {b \over {{a^2}}}\ln |ax + b| + C
\]
\[ \int x {(ax + b)^{ - 2}}dx = {1 \over {{a^2}}}\left[ {\ln |ax + b| + {b \over {ax + b}}} \right] + C
\]
\[ \int {{{dx} \over {x(ax + b)}}} = {1 \over b}\ln |{x \over {ax + b}}| + C
\]
\[ \int {{{(\sqrt {ax + b} )}^n}} dx = {2 \over a}{{{{(\sqrt {ax + b} )}^{n + 2}}} \over {n + 2}} + C \ \ \ \ ,n \ne - 2
\]
\[ \int {{{\sqrt {ax + b} } \over x}} dx = 2\sqrt {ax + b} + b\int {{{dx} \over {x\sqrt {ax + b} }}}
\]
\[ \int {{{dx} \over {x\sqrt {ax + b} }}} = {1 \over {\sqrt b }}\ln \left| {{{\sqrt {ax + b} - \sqrt b } \over {\sqrt {ax + b} + \sqrt b }}} \right| + C
\]
\[ \int {{{dx} \over {x\sqrt {ax - b} }}} = {2 \over {\sqrt b }}{\tan ^{ - 1}}\sqrt {{{ax - b} \over b}} + C
\]
\[ \int {{{\sqrt {ax + b} } \over {{x^2}}}} dx = - {{\sqrt {ax + b} } \over x} + {a \over 2}\int {{{dx} \over {x\sqrt {ax + b} }}} + C
\]
\[ \int {{{dx} \over {{x^2}\sqrt {ax + b} }}} = - {{\sqrt {ax + b} } \over {bx}} - {a \over {2b}}\int {{{dx} \over {x\sqrt {ax + b} }}} + C
\]
1
Thomas' Calculus Early Transcendentals - George B. Thomas Jr., Maurice D. Weir, Joel R. Hass - 13th Edition
نویسنده
علیرضا گلمکانی
شماره کلید
20010
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