What is the integral of sqrt(x^4-2x^3+x^2) ?

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

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integral of sqrt(x^4-2x^3+x^2)

\int x^{4} - 2 x^{3} + x^{2} d x

\frac{x ^{3}}{3} - \frac{x ^{2}}{2} + C

Solution steps

\int x^{4} - 2 x^{3} + x^{2} d x

Factor

x^{4} - 2 x^{3} + x^{2} : x^{2} (x - 1)^{2}

= \int x^{2} (x - 1)^{2} d x

Apply exponent rule:

a^{n} \cdot b^{n} = (a \cdot b)^{n}

x^{2} (x - 1)^{2} = x^{2} (x - 1)^{2}

= \int x^{2} (x - 1)^{2} d x

Apply exponent rule:

a^{n} \cdot b^{n} = (a \cdot b)^{n}

x^{2} (x - 1)^{2} = (x (x - 1))^{2}

= \int (x (x - 1))^{2} d x

(x (x - 1))^{2} = (x (x - 1)),

assuming

x (x - 1) \geq 0

= \int x (x - 1) d x

Expand

x (x - 1) : x^{2} - x

= \int x^{2} - x d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x^{2} d x - \int x d x

\int x^{2} d x = \frac{x ^{3}}{3}

\int x d x = \frac{x ^{2}}{2}

= \frac{x ^{3}}{3} - \frac{x ^{2}}{2}

Add a constant to the solution

= \frac{x ^{3}}{3} - \frac{x ^{2}}{2} + C

f (x) = 4 e^{2 x}

F (x) = (3 x^{6} + 4 x^{3})^{4}

\frac{d y}{d t} = - 4 y + 9 e^{- t}

\int 4 sin^{3} (x) cos^{2} (x) d x

x \to 0 + lim ((1 + x)^{\frac{4}{x}})

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