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integral of (xdx)/(x^2+4)(x^2-5)^0.5
Dear Shivan ,
∫(-x dx) /(x2 + 4)(x2 -5)0.5
substitute x2 -5 = t2 we get xdx =tdt substitute the two substitutions ,
∫-1/(t2+9) dt = - tan-1( t/3) /3 + c
resubstituting the substitution , I = - tan-1( (x2 -5) ^0.5 /3) /3 + c
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