What is the integration of 1 divided by x3 plus x?

Answer 1

4

Answer 2

Note that (1/x³) + x = x-3 + x

Then, by the power rule, we integrate the expression to get:

x-3 + 1/(-3 + 1) + x1 + 1/(1 + 1) + c

= -x-2/2 + x2/2 + c where c is the arbitrary constant

Answer 3

We need to split it using partial fractions, since x^3 + x can be factorised into x(x^2 + 1).

1/(x^3 + x) = A/x + (Bx + C)/(x^2 + 1)

1 = A(x^2 + 1) + x(Bx + C) (eqn 1)

Substituting x = 0 into equation (1),

1 = A + 0

therefore, A = 1.

Substituting x = 1 in to the same equation,

1 = 2A + B + C

Since A = 1,

1 = 2 + B + C

B + C = -1 (eqn 2)

Substituting x = -1,

1 = 2A + B - C

B - C = -1 (eqn 3)

Adding equation 2 and 3 together,

2B = -2

B = -1

-> A = 1, B = -1, C = 0.

Putting these back into our original equation,

1/(x^3 + x) = 1/x - x/(x^2 + 1)

and this can be integrated more easily.

The integral of -

1/x - x/(x^2 + 1) = ln(x) - (1/2) * ln(x^2 + 1)

Therefore, the integral of

1/(x^3 + x) = ln(x) - (1/2) * ln(x^2 + 1)