Integral x^3/(x^2 +1) dx? Do not send if wolfram is used!?

Question

Integral x^3/(x^2 +1) dx? Do not send if wolfram is used!?

Answer 1

Hello,



Please add "Hello", "please" and "Thanks". It DOES help answer your questions.



x³ / (x² + 1)

   = (x³ + x - x) / (x² + 1)

   = (x³ + x)/(x² + 1) - ½ × [2x/(x² + 1)]

   = x - ½ × [2x/(x² + 1)]

   = ½[2x - 2x/(x² + 1)]



Thus,

∫ [x³ / (x² + 1)].dx

   = ∫ ½[2x - 2x/(x² + 1)].dx

   = ½.∫[2x - 2x/(x² + 1)].dx

   = ½.{ ∫ 2x.dx - ∫ [2x/(x² + 1)].dx }

   = ½[x² - ln(x² + 1)]



Regards,

Dragon.Jade :-)

Answer 2

The numerator is a higher order polynomial than the denominator so this can be decomposed by partial fractions. However, u-substitution is also a viable method which ultimately I find to be easier.



∫ x^3/(x^2 + 1) dx --- let u = x^2 + 1, then du/dx = 2x so dx = du/2x, and x^2 = u - 1

= 1/2 ∫ 2x.x^2/(x^2 + 1) dx

= 1/2 ∫ du/dx . (u - 1)/u dx

= 1/2 ∫ (u - 1)/u du

= 1/2 ∫ 1 - 1/u du

= 1/2 [u - ln|u|] + c

= 1/2(x^2 + 1) - 1/2ln(x^2 + 1) + c - the modulus is unnecessary since x^2 + 1 > 0 for all x.



If you're wondering why my answer differs it's because 1/2(x^2 + 1) = 1/2x^2 + 1/2, and the constant 1/2 is absorbed into the arbitrary constant.

Answer 3

For all integrals of the type, f (x)/h (x), where f (x) and h (x) are polynomials of x, first observe the degree of both the polynomials. If f (x) is of the same degree as h (x) or of higher degree, then the first step is the synthetic division and rewriting f (x)/h (x) as q(x) + r(x)/h(x), where q(x) is the quotient polynomial and r(x) is the remainder polynomial where r(x) is of a lesser degree than h (x).

Thus, x^3/(x^2 + 1) can be written as

x - x/(x^2 + 1)

=> ∫ x^3/(x^2 + 1) dx

= ∫ [x - x/(x^2 +1) dx

= x^2/2 - (1/2) ∫ d(x^2 + 1) / (x^2 + 1)

= x^2/2 - (1/2) ln (x^2 + 1) + c.

Answer 4

Divide this expression by the comparing coefficients method:

x³ / (x² + 1) = Ax + B + (Cx + D) / (x² + 1)

x³ = Ax(x² + 1) + B(x² + 1) + Cx + D

x³ = Ax³ + Ax + Bx² + B + Cx + D

x³ = Ax³ + Bx² + (A + C)x + (B + D)

A = 1

B = 0

A + C = 0

C = -A

C = -1

B + D = 0

D = -B

D = 0

x³ / (x² + 1) = x - x / (x² + 1)



Integrate the expression term by term using this result:

∫ x³ / (x² + 1) dx = ∫ [x - x / (x² + 1)] dx

∫ x³ / (x² + 1) dx = ∫ x dx - ∫ 2x / (x² + 1) dx / 2

∫ x³ / (x² + 1) dx = x² / 2 - ln(x² + 1) / 2 + C

Answer 5

x^3/(x^2 +1) = x - x/(x^2 +1) , then it's easy to integrate

Answer 6

Try wolfram alpha👽