What is the integral of (x^3-1)sqrt(x^4-4x+1) ?

The integral of (x^3-1)sqrt(x^4-4x+1) is 1/6 (x^4-4x+1)^{3/2}+C

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integral of (x^3-1)sqrt(x^4-4x+1)

\int (x^{3} - 1) x^{4} - 4 x + 1 d x

\frac{1}{6} (x^{4} - 4 x + 1)^{\frac{3}{2}} + C

Solution steps

\int (x^{3} - 1) x^{4} - 4 x + 1 d x

Apply u-substitution

= \int \frac{u}{4} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int u d u

Apply the Power Rule

= \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = x^{4} - 4 x + 1

= \frac{1}{4} \cdot \frac{2}{3} (x^{4} - 4 x + 1)^{\frac{3}{2}}

Simplify

\frac{1}{4} \cdot \frac{2}{3} (x^{4} - 4 x + 1)^{\frac{3}{2}} : \frac{1}{6} (x^{4} - 4 x + 1)^{\frac{3}{2}}

= \frac{1}{6} (x^{4} - 4 x + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{6} (x^{4} - 4 x + 1)^{\frac{3}{2}} + C

What is the integral of (x^3-1)sqrt(x^4-4x+1) ?
The integral of (x^3-1)sqrt(x^4-4x+1) is 1/6 (x^4-4x+1)^{3/2}+C