What is a if the definite integral of y=(x^3+x) is (2a-1)/4, if the limits of integration are a and a+1?

First, we'll calculate the definite integral, using
Leibniz Newton identity:

Int f(x)dx = F(b) -
F(a)

a = lower limit of
integration

b = upper limit of
integration

Let y = f(x)

Int
(x^3 + x)dx = Int x^3dx + Int xdx

Int x^3dx + Int xdx =
x^4/4 + x^2/2

But Int (x^3 + x)dx = F(a+1) -
F(a)

F(a+1) = (a+1)^4/4 +
(a+1)^2/2

F(a) = a^4/4 +
a^2/2

F(a+1) - F(a) = (a+1)^4/4 + (a+1)^2/2 - a^4/4 -
a^2/2

[(a+1)^4 + 2(a+1)^2 - a^4 - 2a^2]/4 =
(2a-1)/4

[(a+1)^4 + 2(a+1)^2 - a^4 - 2a^2] = 2a -
1

We'll raise the binomials to
powers:

(a^2 + 2a + 1)^2 + 2a^2 + 4a + 2 - a^4 - 2a^2 =
2a-1

a^4 + 4a^2 + 1 + 4a^3 + 2a^2 + 4a + 4a + 2- a^4 =
2a-1

4a^3 + 6a^2 + 6a + 4 =
0

2a^3 + 3a^2 + 3a + 2 =
0

This is an odd reciprocal equation and one of it's roots
is a = -1.

We'll re-write
it:

2a^3 + 3a^2 + 3a + 2 = (a+1)(ca^2 + da+
e)

2a^3 + 3a^2 + 3a + 2 = ca^3 + da^2 + ea + ca^2 + da+
e

c = 2

c + d = 3 => d
= 1

d + e = 3 => e =
2

The reciprocal equation
is:

(a+1)(2a^2 + a+ 2) =
0

We'll calculate the roots of 2a^2 + a+ 2 =
0

delta = 1 - 16 = -15

Since
the discriminant delta is negative, then the equation 2a^2 + a+ 2 = 0 has no real
roots.

The only real root of the resulted
equation is a = -1.