Calculate this integral #int_2^(2sqrt3)1/(x^2sqrt(x^2+4))dx =# ?

2 Answers
Ratnaker Mehta
Jan 30, 2018

# 1/12(3sqrt2-2sqrt3)#.

Explanation:

Let, #I=int_2^(2sqrt3)1/(x^2sqrt(x^2+4))dx#.

We subst. #x=2tany rArr dx=2sec^2ydy#.

Also, when #x=2, 2tany=2rArr y=pi/4, and, #

when #x=2sqrt3 rArr y=pi/3#.

#:. I=int_(pi/4)^(pi/3)(2sec^2y)/{4tan^2ysqrt(4tan^2y+4)}dy#,

#=int_(pi/4)^(pi/3)(2sec^2y)/{4tan^2ysqrt(4tan^2y+4)}dy#,

#=1/4int_(pi/4)^(pi/3)cosy/sin^2ydy#,

#=1/4int_(pi/4)^(pi/3){cosy/siny*1/siny}dy#,

#=1/4int_(pi/4)^(pi/3)cotycscydy#,

#=1/4[-csc y]_(pi/4)^(pi/3)#,

#=-1/4[csc(pi/3)-csc(pi/4)]#,

#=-1/4(2/sqrt3-sqrt2)#,

#=-1/(2sqrt3)+1/4sqrt2#,

#=1/4sqrt2-1/6sqrt3#.

# rArr I=1/12(3sqrt2-2sqrt3)#.

Narad T.
Jan 30, 2018

The answer is #=0.065#

Explanation:

#tan^2x+1=sec^2x#

Perform this integral by substitution

Let #x=2tanu#, #=>#, #dx=2sec^2udu#

Therefore,

#int(dx)/(x^2sqrt(x^2+4))=int(2sec^2 udu)/(4tan^2usqrt(4tan^2u+4))#

#=1/4int(secudu)/tan^2u#

#=1/4int(cosudu)/sin^2u#

Let #v=sinu#, #=>#, #dv=cosudu#

#=1/4int(dv)/v^2#

#=-1/4*1/v#

#=-1/4*1/sinu#

#=-1/4*1/sin(arctan(x/2))#

#=-1/4sqrt(4+x^2)/x+C#

Now, calculate the definite integral

#int_2^(2sqrt3)(dx)/(x^2sqrt(x^2+4))=[-sqrt(4+x^2)/(4x)]_2^(2sqrt3)#

#=(-1/(2sqrt3))-(1/(2sqrt2))#

#=0.065#

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