Integrate: (1+ln x)/square root [(x^x)^2-1]

Answers (1)

Solution: We have ,

                 I=int frac(1+ln x)sqrt(x^x)^2-1dx

             Put ,    (x^x)^2-1=t^2Rightarrow 2(x^x)	imes (x^x)(1+in x)dx=2tdt

                  Rightarrow            (1+ln x)dx=fractdt(t^2+1)

           	herefore         I=int frac1t^2+1dt=	an^-1t+c=	an^-1[sqrt(x^x)^2-1]+c     

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