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

2 Answers
Aug 4, 2016

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

Explanation:

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)
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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))

Aug 4, 2016

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

Explanation:

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))