What is the integral of (sqrt(x)-1)^2(sqrt(x)+1)^2 ?

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

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integral of (sqrt(x)-1)^2(sqrt(x)+1)^2

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

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

Solution steps

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

Multiply by the conjugate of

x - 1 : \frac{x - 1}{x + 1}

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

Simplify

(\frac{x - 1}{x + 1})^{2} (x + 1)^{2} : (x - 1)^{2}

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

Expand

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

= \int x^{2} - 2 x + 1 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 2 x d x + \int 1 d x

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

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

\int 1 d x = x

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

Add a constant to the solution

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

\int 5 - x d x

\frac{d}{d x} (- 3 cot (2 x) sec (2 x))

\int x^{2} + 8 e^{x} + C d x

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{x} + e^{2 x}

\int \frac{1}{3 x + 2} d x

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