What is the integral of ((x-2)^n)/n ?

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

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integral of ((x-2)^n)/n

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int u^{n} d u

= \frac{1}{n} \cdot \int u^{n} d u

Apply the Power Rule

: \frac{u ^{n + 1}}{n + 1}

= \frac{1}{n} \cdot \frac{u ^{n + 1}}{n + 1}

Substitute back

u = x - 2

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{x}{2 x - 7})

\int_{9}^{25} \frac{x - 9}{x} d x

(cos (x) + y sin (x)) d x + (cos (x)) d y = 0

\frac{\partial}{\partial x} (6 x y e^{- x^{2} - y^{2}})

y = 6 sec (x), at x = \frac{π}{6}

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