how does one integrate 1/((1+x^2)^2)?

Answer 1

Integral ( 1/(1 + x^2)^2 dx)

To solve this, use trigonometric substitution.

Let x = tan(t).
dx = sec^2(t) dt

Integral ( 1/(1 + tan^2(t))^2 sec^2(t) dt )

Use the trig identity to convert the denominator to sec^2(t).

Integral ( 1/[sec^2(t)]^2 sec^2(t) dt

Integral ( 1/[sec^4(t)] sec^2(t) dt

This simplifies as

Integral ( 1/[sec^2(t)] dt )

Integral ( cos^2(t) dt)

To integrate this, we need to use the half angle identity,
cos^2(t) = (1/2)(1 + cos(2t)).

Integral ( (1/2) (1 + cos(2t)) dt )

Factor out (1/2).

(1/2) Integral ( (1 + cos(2t)) dt )

Which we can now integrate easily. Remember that the integral of cos(nt) = (1/n) sin(nt), for a constant n.

(1/2) ( t + (1/2)sin(2t)) + C

Distribute the (1/2).

(1/2)t + (1/4)sin(2t) + C

Use the double angle identity sin(2t) = 2sin(t)cos(t)

(1/2)t + (1/4)(2sin(t)cos(t)) + C

(1/2)t + (1/2)sin(t)cos(t) + C

To convert this back in terms of x, remember that

x = tan(t). Therefore,

tan(t) = x/1

By SOHCAHTOA, tan(t) = opp/adj, so

tan(t) = x/1 = opp/adj, so
opp = x,
adj = 1, and by Pythagoras,
hyp = sqrt(1 + x^2)

As a result,
sin(t) = opp/hyp = x/sqrt(1 + x^2)
cos(t) = adj/hyp = 1/sqrt(1 + x^2)

tan(t) = x, so arctan(x) = t. Using these substitution, this:

(1/2)t + (1/2)sin(t)cos(t) + C

becomes this:

(1/2)arctan(x) + (1/2)[x/sqrt(1 + x^2)][1/sqrt(1 + x^2)] + C

This simplifies as

(1/2)arctan(x) + (1/2)[x/(1 + x^2)] + C

Answer 2

Integral Of X 2 1 2

Answer 3

let x = tan z
dx = (1+tanz ^2) dz = (1+x^2) dz
integral 1/(1+x^2)^2 dx
= integral 1/(1+tanz ^2) . dz
= integral cos z ^2 dz
= integral (cos 2z + 1 ) /2 .dz
= sin 2z /4 + z /2
= sin z cos z /2 + z/2
= tan z / (1+tan z^2) /2 + z/2
= x/ [ 2(1+x^2) ] + atan(x) / 2

Answer 4

Use integration by parts.

∫1/((1+x^2)^2) dx
= ∫-1/(2x) d[1/(1+x^2)]
= [-1/(2x)]/(1+x^2) - (1/2)∫[1/(1+x^2)][1/x^2] dx
= [-1/(2x)]/(1+x^2) - (1/2)∫[-1/(1+x^2) + 1/x^2] dx
= [-1/(2x)]/(1+x^2) + (1/2)arctan(x) + 1/(2x)
= (1/2)[x/(1+x^2)] + (1/2)arctan(x)
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Subsitituting x = tan(u) will be a more regular way.

Answer 5

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Answer 6

I only have time to give you the start. Use x = tanA and the identity 1 + (tanA)^2 = (secA)^2. Don't forget to use the substitution in the dx of the integral as well as the function part.

Answer 7

1/((1 + x^2)^2)
expand denomintor brackets:
1/(x^4 + 2x^2 + 1)
rewrite as:
x^-4 + 2x^-2 + 1
Integrate:
-(1/3)x^-3 - 2x^-1 + x + c

There are other ways, but this is the way I'd do it.