Given to solve,

`int sqrt(4+x^2) dx`

using the Trig Substitutions we can solve these type of integrals easily and the solution is as follows

for `sqrt(a+bx^2) ` we can take `x= sqrt(a/b) tan(u)`

so ,For

`int sqrt(4+x^2) dx`

the` x= sqrt(4/1)tan(u)= 2tan(u)`

=> `dx= 2sec^(2) (u) du`

so,

`int sqrt(4+x^2) dx`

= `int sqrt(4+(2tan(u))^2) (2sec^(2) (u) du)`

= `int sqrt(4+4(tan(u))^2) (2sec^(2) (u) du)`

=`int sqrt(4(1+(tan(u))^2)) (2sec^(2) (u) du)`

= `int 2sqrt(1+tan^2(u))(2sec^(2) (u) du)`

= `int 2sec(u)(2sec^(2) (u) du)`

= `int 4sec^(3) (u) du`

`= 4int sec^(3) (u) du`

by applying the Integral Reduction

`int sec^(n) (x) dx`

`= (sec^(n-1) (x) sin(x))/(n-1) + ((n-2)/(n-1)) int sec^(n-2) (x) dx`

so ,

`4int sec^(3) (u) du`

= `4[(sec^(3-1) (u) sin(u))/(3-1) + ((3-2)/(3-1)) int sec^(3-2) (u)du]`

= `4[(sec^(2) (u) sin(u))/(2) + ((1)/(2)) int sec (u)du]`

=`4[(sec^(2) (u) sin(u))/(2) + (1/2) (ln(tan(u)+sec(u)))]`

but` x= 2tan(u)`

=>` x/2 = tan(u)`

`u = tan^(-1) (x/2)`

so,

`4[(sec^(2) (u) sin(u))/(2) + (1/2) (ln(tan(u)+sec(u)))]`

`=4[(sec^(2) ( tan^(-1) (x/2)) sin( tan^(-1) (x/2)))/(2) + (1/2) (ln(tan( tan^(-1) (x/2))+sec( tan^(-1) (x/2))))]`

=`4[(sec^(2) ( tan^(-1) (x/2)) sin( tan^(-1) (x/2)))/(2) + (1/2) (ln((x/2))+sec( tan^(-1) (x/2)))] +c`

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