What is the integral of ((x+1))/(\sqrt[4]{4)}(x-1) ?

The integral of ((x+1))/(\sqrt[4]{4)}(x-1) is 1/(sqrt(2))((x^3)/3-x)+C

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integral of ((x+1))/(\sqrt[4]{4)}(x-1)

\int \frac{( x + 1 )}{4 4} (x - 1) d x

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

Solution steps

\int \frac{x + 1}{4 4} (x - 1) d x

Take the constant out:

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

= \frac{1}{4 4} \cdot \int (x + 1) (x - 1) d x

44 = 2

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

Expand

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

= \frac{1}{2} \cdot \int x^{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

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

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

\int 1 d x = x

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

Add a constant to the solution

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

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