Question

`int_(-1)^1x^2/(sqrt(x^2+1)+x+1)dx` = ?

Find `int_(-1)^1x^2/(sqrt(x^2+1)+x+1)dx`

Answer 1

`int_(-1)^1 x^2/(sqrt(x^2+1)+x+1)*dx=1/3`

Explanation:

`int_(-1)^1 x^2/(sqrt(x^2+1)+x+1)*dx`

=`int_(-1)^1 (x^2*(x+1-sqrt(x^2+1))*dx)/((x+1)^2-(x^2+1))`

=`int_(-1)^1 (x^2*(x+1-sqrt(x^2+1))*dx)/(2x)`

=`1/2int_(-1)^1 x*(x+1-sqrt(x^2+1))*dx`

=`1/2int_(-1)^1 (x^2+x-xsqrt(x^2+1))*dx`

=`[1/6x^3+1/4x^2-1/6(x^2+1)^(3/2)]_(-1)^1`

=`1/3`

Answer 2

`int_(-1)^1 x^2/(sqrt(x^2+1)+x+1)*dx=1/3`

Explanation:

`int_(-1)^1 x^2/(sqrt(x^2+1)+x+1)*dx`

After using `x=tany` and `dx=(secy)^2*dy` transforms, this integral became

`int_(-pi/4)^(pi/4) (tany)^2/(secy+tany+1)*(secy)^2*dy`

=`int_(-pi/4)^(pi/4) ((tany)^2*(tany+1-secy)*(secy)^2*dy)/((tany+1)^2-(secy)^2)`

=`int_(-pi/4)^(pi/4) ((tany)^2*(tany+1-secy)*(secy)^2*dy)/((tany)^2+1+2tany-(secy)^2)`

=`int_(-pi/4)^(pi/4) ((tany)^2*(tany+1-secy)*(secy)^2*dy)/((secy)^2+2tany-(secy)^2)`

=`int_(-pi/4)^(pi/4) ((tany)^2*(tany+1-secy)*(secy)^2*dy)/(2tany)`

=`1/2int_(-pi/4)^(pi/4) tany*(tany+1-secy)*(secy)^2*dy`

=`1/2int_(-pi/4)^(pi/4) (tany)^2*(secy)^2*dy`+`1/2int_(-pi/4)^(pi/4) (tany)*(secy)^2*dy`-`1/2int_(-pi/4)^(pi/4) (secy)^3*tany*dy`

=`[1/6(tany)^3+1/4(tany)^2-1/6(secy)^3]_(-pi/4)^(pi/4)`

=`1/3`