Question

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

Answer 1

`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)`.

Answer 2

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`