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`int xarcsec(x^2+1) dx` Use integration tables to find the indefinite integral.

Indefinite integral are written in the form of `int f(x) dx = F(x) +C`


 where: `f(x)` as the integrand


           `F(x)` as the anti-derivative function 


           `C`  as the arbitrary constant known as constant of integration


For the given problem `int xarcsec(x^2+1) dx,` it has a integrand in a form of  inverse secant function. The integral resembles one of the formulas from the integration as :  `int arcsec (u/a)du = u*arcsin(u/a) +-aln(u+sqrt(u^2-a^2))+C` .


where we use: `(+)`  if `0ltarcsec (u/a)ltpi/2`


                    `(-)` if `pi/2ltarcsec(u/a)ltpi`


Selecting the sign between `(+)` and` (-) ` will be crucial when solving for definite integral with given boundary values `[a,b]` .


 For easier comparison, we may apply u-substitution by letting:


`u =x^2+1` then `du = 2x dx ` or `(du)/2`


Plug-in the values `int xarcsec(x^2+1) dx` , we get:


`int xarcsec(x^2+1) dx=int arcsec(x^2+1) * xdx`


                                        `= int arcsec(u) * (du)/2`


Apply the basic properties of integration: `int c*f(x) dx= c int f(x) dx` .


`int arcsec(u) * (du)/2= 1/2int arcsec(u) du`


                         or `1/2 int arcsec(u/1) du`


Applying the aforementioned formula from the integration table, we get:


`1/2 int arcsec(u/1) du=1/2 *[u*arcsin(u/1) +-1ln(u+sqrt(u^2-1^2))]+C`


                                `=1/2 *[u*arcsin(u) +-ln(u+sqrt(u^2-1))]+C`


                                `=(u*arcsin(u))/2 +-(ln(u+sqrt(u^2-1)))/2+C`


Plug-in `u =x^2+1` on `(u*arcsin(u))/2 +-(ln(u+sqrt(u^2-1)))/2+C` , we get the indefinite integral as:


`int xarcsec(x^2+1) dx=((x^2+1)*arcsin(x^2+1))/2 +-(ln(x^2+1+sqrt((x^2+1)^2-1)))/2+C`


`=(x^2arcsin(x^2+1))/2+arcsin(x^2+1)/2 +-ln(x^2+1+sqrt(x^4+2x^2))/2+C`


`=(x^2arcsin(x^2+1))/2+arcsin(x^2+1)/2 +-ln(x^2+1+sqrt(x^2(x^2+2)))/2+C`


`=(x^2arcsin(x^2+1))/2+arcsin(x^2+1)/2 +-ln(x^2+1+|x|sqrt(x^2+2))/2+C`

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