integral of ( xdx)/(x^2+4)(x^2-5)^0.5

integral of (xdx)/(x^2+4)(x^2-5)^0.5


1 Answers

AJIT AskiitiansExpert-IITD
68 Points
12 years ago

Dear Shivan ,

 ∫(-x dx) /(x2 + 4)(x2 -5)0.5  

substitute x2 -5 = t  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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