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

The integral of 2(1+sqrt(x-1))^3 is 3x^2+4/5 x(x-1)^{3/2}-4x+16/5 (x-1)^{3/2}+C

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integral of 2(1+sqrt(x-1))^3

\int 2 (1 + x - 1)^{3} d x

3 x^{2} + \frac{4}{5} x (x - 1)^{\frac{3}{2}} - 4 x + \frac{16}{5} (x - 1)^{\frac{3}{2}} + C

Solution steps

\int 2 (1 + x - 1)^{3} d x

Take the constant out:

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

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

Expand

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

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

= 2 (\int 3 x d x + \int 3 x - 1 d x + \int (x - 1)^{\frac{3}{2}} d x - \int 2 d x)

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

\int 3 x - 1 d x = 2 (x - 1)^{\frac{3}{2}}

\int (x - 1)^{\frac{3}{2}} d x = \frac{2}{5} x (x - 1)^{\frac{3}{2}} - \frac{2}{5} (x - 1)^{\frac{3}{2}}

\int 2 d x = 2 x

= 2 (\frac{3 x ^{2}}{2} + 2 (x - 1)^{\frac{3}{2}} + \frac{2}{5} x (x - 1)^{\frac{3}{2}} - \frac{2}{5} (x - 1)^{\frac{3}{2}} - 2 x)

Simplify

2 (\frac{3 x ^{2}}{2} + 2 (x - 1)^{\frac{3}{2}} + \frac{2}{5} x (x - 1)^{\frac{3}{2}} - \frac{2}{5} (x - 1)^{\frac{3}{2}} - 2 x) : 3 x^{2} + \frac{4}{5} x (x - 1)^{\frac{3}{2}} - 4 x + \frac{16}{5} (x - 1)^{\frac{3}{2}}

= 3 x^{2} + \frac{4}{5} x (x - 1)^{\frac{3}{2}} - 4 x + \frac{16}{5} (x - 1)^{\frac{3}{2}}

Add a constant to the solution

= 3 x^{2} + \frac{4}{5} x (x - 1)^{\frac{3}{2}} - 4 x + \frac{16}{5} (x - 1)^{\frac{3}{2}} + C

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What is the integral of 2(1+sqrt(x-1))^3 ?
The integral of 2(1+sqrt(x-1))^3 is 3x^2+4/5 x(x-1)^{3/2}-4x+16/5 (x-1)^{3/2}+C