Question #2250e - Socratic

1/2"arcsec"^2 (x)+C >I=int ("arcsec" (x))/ (xsqrt (x^2-1))dx Let x=sectheta. This implies that dx=secthetatanthetad theta. Also it means that theta="arcsec" (x). Then: I=inttheta/ (secthetasqrt (sec^2theta-1)) (secthetatanthetad theta) Note that sec^2theta-1=tan^2theta: I=inttheta/ (secthetatantheta) (secthetatantheta)d theta I=intthetacolor (white).d theta I=1/2theta^2+C I=1/2"arcsec"^2 (x)+C

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