Calculate the definite integral of y=x^3/square root(x^4+1). limits of integration: x=0 and x=1

We'll apply Leibniz-Newton formula to evaluate the
definite integral:

Int f(x)dx = F(b) - F(a), where a and b
are the limits of integration.

First, we'll determine the
indefinite integral. We'll change the variable as method of solving the
integral.

Int x^3dx/sqrt(x^4 +
1)

We notice that if we'll put x^4 + 1 = t and we'll
differentiate, we'll get the numerator.

4x^3dx =
dt

x^3dx = dt/4

We'll re-write
the integral:

Int x^3dx/sqrt(x^4 + 1) = Int (dt/4)/sqrt
t

Int (dt/4)/sqrt t = (1/4)*Int dt/sqrt
t

(1/4)*Int dt/sqrt t = (1/4)*[t^(-1/2 + 1)/(-1/2 + 1)] +
C

(1/4)*Int dt/sqrt t = (1/2)*sqrt t +
C

Int f(x)dx = (1/2)*sqrt (x^4 + 1) +
C

We'll evaluate the definite
integral:

Int f(x)dx = (1/2)*sqrt (1^4 + 1) - (1/2)*sqrt
(0^4 + 1)

Int f(x)dx = (1/2)*(sqrt2 -
1)

The definite integral of the function f(x)
= x^3/sqrt(x^4 + 1), is Int f(x)dx = (1/2)*(sqrt2 -
1).