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How do you integrate #int 1/sqrt(4x^2+16x-5) # using trigonometric substitution?

1 Answer

Jun 3, 2018

#-1/8*ln|t-4|+C#
where #t=sqrt(4x^2+16x-5)-2x#

Explanation:

Setting
#sqrt(4x^2+16x-5)=2x+t#
then we get (by squaring)

#x=(5+t^2)/(16-4t)#
#dx=(-4t^2+32t+20)/(16-4t)^2dt#
so

#sqrt(4x^2+16x-5)=2(5+t^2)/(16-4t)+t=(-t^2+8t+5)/(8-2t)#
and our integral will be

#int (8-2t)/(-t^2+8t+5)*4(-t^2+8t+5)/(16-4t)^2dt#

#-1/8int dt/(t-4)=-1/8*ln|t-4|+C#

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