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

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

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integral of (sqrt(x)-1)^2(1+x^{1/2})^2

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

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

Solution steps

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

Expand

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

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

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

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

\int 2 x d x = \frac{4}{3} x^{\frac{3}{2}}

\int 1 d x = x

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

Add similar elements:

- \frac{4}{3} x^{\frac{3}{2}} + \frac{4}{3} x^{\frac{3}{2}} = 0

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

Add a constant to the solution

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

arctan (16 x)

y^{^{'}} = \frac{y + x ^{2} + y ^{2}}{x}

y = 2 x^{3} + 3 x^{2} - 12 x + 6

\int \frac{1}{9 x ^{2}} d x

\frac{\partial}{\partial x} (x^{2} y z^{3} (x - 10 y))

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