What is the integral of 4(x-1)in(x-1) ?

The integral of 4(x-1)in(x-1) is 4in((x^3)/3-x^2+x)+C

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integral of 4(x-1)in(x-1)

\int 4 (x - 1) in (x - 1) d x

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

Solution steps

\int 4 (x - 1) in (x - 1) d x

Take the constant out:

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

= 4 in \cdot \int (x - 1) (x - 1) d x

Simplify

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

= 4 in \cdot \int (x - 1)^{2} d x

Expand

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

= 4 in \cdot \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

= 4 in (\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

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

Add a constant to the solution

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

f (x) = \frac{x cos ( π x ) + 1}{x - 1}

(3 x + 13)^{\frac{1}{2}}

\frac{d}{d x} (f (x) (2 x))

\int_{- 1}^{1} e^{x} - x^{2} + 6 d x

x \to 5 lim (\frac{2 - 25}{x - 5})

What is the integral of 4(x-1)in(x-1) ?
The integral of 4(x-1)in(x-1) is 4in((x^3)/3-x^2+x)+C