\( \int {({1 \over x} - {2 \over {\sqrt {1 - {x^2}} }})dx} = \)( )
A: \( \ln \left| x \right| + 2\arcsin x + C \)
B: \( \ln \left| x \right| - 2\arcsin x + C \)
C: \(- \ln \left| x \right| - 2\arcsin x + C \)
D: \(- \ln \left| x \right| +2\arcsin x + C \)
A: \( \ln \left| x \right| + 2\arcsin x + C \)
B: \( \ln \left| x \right| - 2\arcsin x + C \)
C: \(- \ln \left| x \right| - 2\arcsin x + C \)
D: \(- \ln \left| x \right| +2\arcsin x + C \)
举一反三
- 函数\(y = {\left( {\arcsin x} \right)^2}\)的导数为（ ）. A: \(2\arcsin x{1 \over {\sqrt {1 - {x^2}} }}\) B: \( - 2\arcsin x{1 \over {\sqrt {1 - {x^2}} }}\) C: \(2\arcsin x{1 \over {\sqrt {1 + {x^2}} }}\) D: \( - 2\arcsin x{1 \over {\sqrt {1 + {x^2}} }}\)
- $\int {{1 \over {3 + 5\cos x}}} dx = \left( {} \right)$ A: ${1 \over 4}\ln \left| {{{2\cos x + \sin x} \over {2\cos x - \sin x}}} \right| + C$ B: ${1 \over 4}\ln \left| {{{2\cos {x \over 2} + \sin {x \over 2}} \over {2\cos {x \over 2} - \sin {x \over 2}}}} \right| + C$ C: $\ln \left| {{{\cos {x \over 2} + \sin {x \over 2}} \over {\cos {x \over 2} - \sin {x \over 2}}}} \right| + C$ D: $\ln \left| {{{\cos x + \sin x} \over {\cos x - \sin x}}} \right| + C$
- 函数\(y = 1{\rm{ + }}{1 \over x}\)的导数为（ ）. A: \({\rm{ - }}{1 \over { { x^2}}}\) B: \({1 \over { { x^2}}}\) C: \(\ln \left| x \right|\) D: \( - \ln \left| x \right|\)
- 若连续函数\(f\left( x \right)\)满足关系式\(f\left( x \right) = \int_0^{2x} {f\left( { { t \over 2}} \right)} \,dt + \ln 2\)，则\(f\left( x \right)\)等于（ ）。 A: \({e^{2x}}\ln 2\) B: \({e^x}\ln 2\) C: \({e^x} + \ln 2\) D: \({e^{2x}} + \ln 2\)
- 函数\(z = {\left( {xy} \right)^x}\)的全微分为 A: \(dz = \left( { { {\left( {xy} \right)}^x} + \ln xy} \right)dx + x{\left( {xy} \right)^x}dy\) B: \(dz = \left( { { {\left( {xy} \right)}^x} + \ln xy} \right)dx + { { x { { \left( {xy} \right)}^x}} \over y}dy\) C: \(dz = {\left( {xy} \right)^x}\ln xydx + { { x { { \left( {xy} \right)}^x}} \over y}dy\) D: \(dz = {\left( {xy} \right)^x}\left( {1 + \ln xy} \right)dx + { { x { { \left( {xy} \right)}^x}} \over y}dy\)