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

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

2 Answers
Apr 3, 2018

#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#

Apr 3, 2018

#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#