What is the integral of (1/(x^2))/((a-1/x)^k) ?

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

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integral of (1/(x^2))/((a-1/x)^k)

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

\frac{( a x - 1 ) ^{- k + 1}}{x ^{- k + 1} ( - k + 1 )} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{k}} d u

= \int \frac{1}{u ^{k}} d u

Apply the Power Rule

: \frac{u ^{- k + 1}}{- k + 1}

= \frac{u ^{- k + 1}}{- k + 1}

Substitute back

u = a - \frac{1}{x}

= \frac{( a - \frac{1}{x} ) ^{- k + 1}}{- k + 1}

Simplify

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

= \frac{( a x - 1 ) ^{- k + 1}}{x ^{- k + 1} ( - k + 1 )}

Add a constant to the solution

= \frac{( a x - 1 ) ^{- k + 1}}{x ^{- k + 1} ( - k + 1 )} + C

\frac{\partial}{\partial t} (t e^{- 2 x} + cos (x))

f (x) = x^{2}, at x = 1.9

\int \frac{1 - 0.4 x}{( 1 - x ) ( 5 - x )} d x

\int_{- \infty}^{\infty} t d t

\frac{d}{d x} ((5 - (4 - 3 x)^{2})^{3})

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