How do you find the antiderivative of int 1/(x^2(1+x^2)) dx∫1x2(1+x2)dx?

2 Answers
Douglas K.
Oct 16, 2016

Partial fraction expansion gives you two trivial integrals:

int1/(x^2(x^2 + 1))dx = int1/x^2dx - int1/(x^2 + 1)dx∫1x2(x2+1)dx=∫1x2dx−∫1x2+1dx

Explanation:

Use partial fraction expansion:

1/(x^2(x^2 + 1)) = A/x + B/x^2 + (Cx +D)/(x^2 + 1)1x2(x2+1)=Ax+Bx2+Cx+Dx2+1

1 = A(x(x^2 + 1)) + B(x^2 + 1) + (Cx +D)(x^2)1=A(x(x2+1))+B(x2+1)+(Cx+D)(x2)

Let x = 0x=0:

B = 1

1 - (x^2 + 1) = A(x(x^2 + 1)) + (Cx +D)(x^2)1−(x2+1)=A(x(x2+1))+(Cx+D)(x2)

Let x = 1:

-1 = A(1(1^2 + 1)) + (C + D)(1^2)−1=A(1(12+1))+(C+D)(12)

-1 = 2A + C + D

Let x = -1

1 - (-1^2 + 1) = A(-1(-1^2 + 1)) + (C-1 +D)(-1^2)1−(−12+1)=A(−1(−12+1))+(C−1+D)(−12)

-1 = -2A - C + D

D = -1

A = C = 0

Check:

1/x^2 - 1/(x^2 + 1) = 1x2−1x2+1=

1/x^2(x^2 + 1)/(x^2 + 1) - 1/(x^2 + 1)(x^2)/(x^2) = 1x2x2+1x2+1−1x2+1x2x2=

(x^2 + 1)/(x^2(x^2 + 1)) - (x^2)/(x^2(x^2 + 1)) = x2+1x2(x2+1)−x2x2(x2+1)=

1/(x^2(x^2 + 1)) 1x2(x2+1) This checks.

int1/(x^2(x^2 + 1))dx = int1/x^2dx - int1/(x^2 + 1)dx∫1x2(x2+1)dx=∫1x2dx−∫1x2+1dx

mason m
Oct 21, 2016

-1/x-arctanx+C−1x−arctanx+C

Explanation:

This method avoids partial fractions and uses a trig substitution.

I=int1/(x^2(1+x^2))dxI=∫1x2(1+x2)dx

Let x=tanthetax=tanθ. This implies that dx=sec^2thetad thetadx=sec2θdθ. Thus:

I=intsec^2theta/(tan^2theta(1+tan^2theta))d thetaI=∫sec2θtan2θ(1+tan2θ)dθ

Since 1+tan^2theta=sec^2theta1+tan2θ=sec2θ:

I=int1/tan^2thetad theta=intcot^2thetad thetaI=∫1tan2θdθ=∫cot2θdθ

Note that cot^2theta=csc^2theta-1cot2θ=csc2θ−1:

I=intcsc^2thetad theta-intd thetaI=∫csc2θdθ−∫dθ

These are common integrals:

I=-cottheta-theta+CI=−cotθ−θ+C

Note that tantheta=xtanθ=x, so cottheta=1/xcotθ=1x and theta=arctanxθ=arctanx.

I=-1/x-arctanx+CI=−1x−arctanx+C

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