Given to solve,

`int sqrt(9+16x^2) dx`

by using the trig substitution , we can solve the integral

for `sqrt(a +bx^2)` ` dx ` the `x` is given as

`x= sqrt(a/b) tan(u)`

so,

for the integral

`int sqrt(9+16x^2) dx`

let` x=sqrt(9/16) tan(u) = (3/4) tan(u)`

=>` dx = (3/4) sec^2(u) du`

so,

`int sqrt(9+16x^2) dx`

=`int [sqrt(9(1+16/9 x^2))] ((3/4) sec^2(u) du)`

= `3 int [sqrt(1+(16/9)x^2)] ((3/4) sec^2(u) du)`

= `3 int sqrt(1+(16/9)((3/4) tan(u))^2) ((3/4) sec^2(u) du)`

= `3 int [sqrt(1+(16/9)(9/16)(tan^2 u))] ((3/4) sec^2(u) du)`

= `(9/4) int sqrt(1+tan^2 u) (sec^2(u) du)`

= `(9/4) int sqrt(sec^2 u) (sec^2(u) du)`

`= (9/4) int sec u (sec^2(u) du)`

`= (9/4) int (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 ,

`(9/4)int sec^(3) (u) du`

= `(9/4)[(sec^(3-1) (u) sin(u))/(3-1) + ((3-2)/(3-1)) int sec^(3-2) (u)du]`

= `(9/4)[(sec^(2) (u) sin(u))/(2) + ((1)/(2)) int sec (u)du]`

=`(9/4)[(sec^(2) (u) sin(u))/(2) + (1/2) (ln(tan(u)+sec(u)))]`

but we know

`x= (3/4) tan(u)`

= > `4x/3 = tan(u)`

=> `u =arctan(4x/3)`

so,

=`(9/4)[(sec^(2) (u) sin(u))/(2) + (1/2) (ln(tan(u)+sec(u)))]`

=`(9/4)[(sec^(2) (arctan(4x/3)) sin(arctan(4x/3)))/(2) + (1/2) (ln(tan(arctan(4x/3))+sec(arctan(4x/3))))]`

=`(9/4)[(sec^(2) (arctan(4x/3)) sin(arctan(4x/3)))/(2) + (1/2) (ln((4x/3))+sec(arctan(4x/3)))]+c`