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

The integral of (sqrt(x)-1)^3 is 2/5 x^{5/2}-(3x^2)/2+2x^{3/2}-x+C

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

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

\frac{2}{5} x^{\frac{5}{2}} - \frac{3 x ^{2}}{2} + 2 x^{\frac{3}{2}} - x + C

Solution steps

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

Apply u-substitution

= \int (u - 1)^{3} \cdot 2 u d u

Take the constant out:

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

= 2 \cdot \int (u - 1)^{3} u d u

Expand

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

= 2 \cdot \int u^{4} - 3 u^{3} + 3 u^{2} - u d u

Apply the Sum Rule:

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

= 2 (\int u^{4} d u - \int 3 u^{3} d u + \int 3 u^{2} d u - \int u d u)

\int u^{4} d u = \frac{u ^{5}}{5}

\int 3 u^{3} d u = \frac{3 u ^{4}}{4}

\int 3 u^{2} d u = u^{3}

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

= 2 (\frac{u ^{5}}{5} - \frac{3 u ^{4}}{4} + u^{3} - \frac{u ^{2}}{2})

Substitute back

u = x

= 2 (\frac{( x ) ^{5}}{5} - \frac{3 ( x ) ^{4}}{4} + (x)^{3} - \frac{( x ) ^{2}}{2})

Simplify

2 (\frac{( x ) ^{5}}{5} - \frac{3 ( x ) ^{4}}{4} + (x)^{3} - \frac{( x ) ^{2}}{2}) : \frac{2}{5} x^{\frac{5}{2}} - \frac{3 x ^{2}}{2} + 2 x^{\frac{3}{2}} - x

= \frac{2}{5} x^{\frac{5}{2}} - \frac{3 x ^{2}}{2} + 2 x^{\frac{3}{2}} - x

Add a constant to the solution

= \frac{2}{5} x^{\frac{5}{2}} - \frac{3 x ^{2}}{2} + 2 x^{\frac{3}{2}} - x + C

What is the integral of (sqrt(x)-1)^3 ?
The integral of (sqrt(x)-1)^3 is 2/5 x^{5/2}-(3x^2)/2+2x^{3/2}-x+C