ดูหนึ่งข้อความ
  #6  
Old 11 เมษายน 2009, 22:37
V.Rattanapon V.Rattanapon ไม่อยู่ในระบบ
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1.\[
\int {\sqrt {10^{3x} } dx = \int {\left( {10^{\frac{3}{2}} } \right)} } ^x dx = \frac{{2\sqrt {10^{3x} } }}{{3\ln 10}} + c
\]
2.\[
\int {a^x b^x dx = \int {\left( {ab} \right)} } ^x dx = \frac{{a^x b^x }}{{\ln \left( {ab} \right)}} + c
\]
3.\[
\int {\frac{{x + 2}}{{x^2 + 1}}dx = } \int {\frac{x}{{x^2 + 1}}dx + \int {\frac{2}{{x^2 + 1}}dx = } } \frac{1}{2}\int {\frac{{d\left( {x^2 + 1} \right)}}{{x^2 + 1}} + \int {\frac{2}{{x^2 + 1}}dx = } \frac{1}{2}\ln } \left( {x^2 + 1} \right) + 2\arctan x + c
\]
4.\[
\int {\frac{{3dx}}{{x^2 + 4x + 5}} = 3\int {\frac{{dx}}{{\left( {x + 2} \right)^2 + 1}} = } } 3\int {\frac{{d\left( {x + 2} \right)}}{{\left( {x + 2} \right)^2 + 1}} = } 3\arctan \left( {x + 2} \right) + c
\]
5.\[
\int {\frac{{\left( {1 + \tan ^2 x} \right)}}{{1 + \tan x}}} dx = \int {\frac{{\sec ^2 x}}{{1 + \tan x}}} dx\int {\frac{{d\left( {1 + \tan x} \right)}}{{1 + \tan x}} = \ln \left| {1 + \tan x} \right| + c}
\]
6.\[
\int {\frac{{\tan x}}{{1 - \tan ^2 x}}dx = \int {\frac{{\sin x\cos x}}{{\cos ^2 x - \sin ^2 x}}dx = } } \frac{1}{2}\int {\frac{{\sin 2x}}{{\cos 2x}}} dx = \frac{1}{4}\int {\tan 2xd\left( {2x} \right) = \frac{1}{4}\ln \left| {\sec 2x} \right| + c}
\]
7.\[
\int {\frac{{1 + \sqrt {x + 1} }}{{\sqrt {x + 1} }}} dx = \int {\left( {\frac{1}{{\sqrt {x + 1} }} + 1} \right)dx = 2\sqrt {x + 1} + x + c}
\]
8.\[
\int {\frac{{\sin 2x}}{{\cos x}}dx = 2\int {\sin xdx = - 2\cos x + c} }
\]
9.\[
\int {\frac{{x^4 - x}}{{x^3 }}dx = \int {\left( {x - \frac{1}{{x^2 }}} \right)} } dx = \frac{{x^2 }}{2} + \frac{1}{x} + c
\]
10.\[
\int {\frac{{e^x }}{{\sqrt {1 + e^{2x} } }}dx = } \int {\frac{{d\left( {e^x } \right)}}{{\sqrt {1 + e^{2x} } }} = } \ln \left( {\sqrt {1 + e^{2x} } + e^x } \right) + c
\]

11 เมษายน 2009 22:39 : ข้อความนี้ถูกแก้ไขแล้ว 1 ครั้ง, ครั้งล่าสุดโดยคุณ V.Rattanapon
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