What is the antiderivative of #(2x)/(sqrt(5 + 4x))#?

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Answer 1 · 2016-08-23T12:36:18

#I=1/8{5+4x-10lnsqrt(5+4x)}+C#, or,

#I=1/4(2x-5lnsqrt(5+4x))+K#, where, #K=5/8+C#.

We take substn. #5+4x=t^2, so that, 4dx=2tdt, i.e., dx=1/2dt#.

Also, #5+4x=t^2 rArr x=(t^2-5)/4#. Therefore,

#I=int(2x)/sqrt(5+4x)dx=1/2int{2(t^2-5)/4}/sqrt(t^2)dt=1/4int(t^2-5)/tdt#

#=1/4int(t^2/t-5/t)dt=1/4(t^2/2-5ln|t|)#

#=1/8(t^2-10ln|t|)#. Therefore,

#I=1/8{5+4x-10lnsqrt(5+4x)}+C#

Readjusting the Constant of Integration, we may like,

#I=1/4(2x-5lnsqrt(5+4x))+K#, where, #K=5/8+C#.

Enjoy Maths.!