How do you integrate #int dx/(x^2+1)^2# using trig substitutions?

Question

15284 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-04T11:29:44

#int dx/(x^2+1)^2=(1/2) (tan^-1(x)+x/(1+x^2))#

#int dx/(x^2+1)^2 #

Use #x=tan(a)#

#dx=sec^2(a)da#

#intdx/(x^2+1)^2=int (sec^2(a)da)/(1+tan^2a)^2#

Use the identity # 1+tan^2(a)=sec^2(a)#

#intdx/(x^2+1)^2=int (sec^2(a)da)/sec^4(a)#

#=int (da)/sec^2(a)#

#=int cos^2(a) da#
#=int ((1+cos(2a))/2 )da#

#=(1/2)(int (da)+int cos(2a)da)#
#=(1/2)(a+sin(2a)/2)#
#=(1/2)(a+(2sin(a)cos(a))/2)#
#=(1/2) (a+sin(a).cos(a))#

we know that #a=tan^-1(x)#
enter image source here
#sin(a)=x/(sqrt(1+x^2)#
#cos(a)=x/(sqrt(1+x^2#

#int dx/(x^2+1)^2= (1/2) (tan^-1(x) +sin(sin^-1(x/(sqrt(1+x^2)))cos(cos^-1(1/(sqrt(1+x^2))))#
#=(1/2) (tan^-1(x)+(x/(sqrt(1+x^2))1/sqrt(1+x^2))#

#=(1/2) (tan^-1(x)+x/(1+x^2))#

Answer 2 · 2016-08-04T11:32:41

#int dx/(x^2+1)^2 = 1/2(arctan(x) + x/(x^2+1))#

#int dx/(x^2+1)^2# performing the substitution

#x = tan(y)# and consequently
#dx = dy/(cos(y)^2)#

we have

#int dx/(x^2+1)^2 equiv int dy/(cos(y)^2 (1/cos(y)^4)) = int cos(y)^2dy#

but

#d/(dy)(sin (y) cos (y)) = cos(y)^2-sin(y)^2 = 2 cos(y)^2-1#

then

#int cos(y)^2 dy = 1/2(y + sin(y) cos(y))#

Finally, recalling #y = arctan(x)# we have

#int dx/(x^2+1)^2 = 1/2(arctan(x) + x/(x^2+1))#