What is the integral of (a-bx)^{-3} ?

The integral of (a-bx)^{-3} is 1/(2b(a-bx)^2)+C

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integral of (a-bx)^{-3}

\int (a - b x)^{- 3} d x

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

Solution steps

\int (a - b x)^{- 3} d x

Apply u-substitution

: \int - \frac{1}{b u ^{3}} d u

= \int - \frac{1}{b u ^{3}} d u

Take the constant out:

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

= - \frac{1}{b} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

= - \frac{1}{b} (- \frac{1}{2 u ^{2}})

Substitute back

u = a - b x

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

Simplify

- \frac{1}{b} (- \frac{1}{2 ( a - b x ) ^{2}}) : \frac{1}{2 b ( a - b x ) ^{2}}

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

Add a constant to the solution

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

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What is the integral of (a-bx)^{-3} ?
The integral of (a-bx)^{-3} is 1/(2b(a-bx)^2)+C