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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