# 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)...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

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`

Approved by eNotes Editorial Team