91Ó°ÊÓ

First, let's rewrite the provided integral: $$\int_{0}^{1} \frac{x^{3}}{x^{4}-1} dx$$

The denominator of the given function can be factored as \((x^4 - 1) = (x^2+1)(x^2-1) = (x^2+1)(x+1)(x-1)\). Using partial fractions decomposition, we write: $$\frac{x^{3}}{x^{4}-1} = \frac{Ax^2 + B}{x^2 + 1} + \frac{C}{x+1} + \frac{D}{x-1}$$ Now, clear the denominators by multiplying both sides by \((x^4-1)\): $$x^3 = A(x+1)(x-1)(x^2+1) + C(x^2+1)(x^2-1) + D(x+1)(x^2+1)(x-1)$$ Now, we can find the values of A, B, C and D by substituting the following values into the above equation: 1. \(x=-1\) gives \(-A = 0 \Rightarrow A = 0\) 2. \(x=1\) gives \(D = 0 \Rightarrow D = 0\) 3. \(x=0\) gives \(-C = 0 \Rightarrow C = 0\) As A, C and D are all zero, the partial fractions decomposition simplifies to: $$\frac{x^{3}}{x^{4}-1} = \frac{B}{x^2 + 1}$$ Comparing the coefficients of the powers of \(x\), we find that \(B=1\).

Now, we can find the integral of the above partial fraction decomposition: $$\int_{0}^{1} \frac{x^{3}}{x^{4}-1} dx = \int_{0}^{1} \frac{1}{x^2 + 1} dx$$ Here, we will notice the given integral represents the arctangent function: $$\int_{0}^{1} \frac{1}{x^2 + 1} dx = \left[\arctan(x)\right]_0^1$$

Now we can find the definitive integral by evaluating the limits: $$\left[\arctan(x)\right]_0^1 = \arctan(1) - \arctan(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$$ The definite integral evaluates to \(\frac{\pi}{4}\).