What is the integral of (72x)/((x-7)^{10)} ?

The integral of (72x)/((x-7)^{10)} is 72(-1/(8(x-7)^8)-7/(9(x-7)^9))+C

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integral of (72x)/((x-7)^{10)}

\int \frac{72 x}{( x - 7 ) ^{10}} d x

72 (- \frac{1}{8 ( x - 7 ) ^{8}} - \frac{7}{9 ( x - 7 ) ^{9}}) + C

Solution steps

\int \frac{72 x}{( x - 7 ) ^{10}} d x

Take the constant out:

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

= 72 \cdot \int \frac{x}{( x - 7 ) ^{10}} d x

Apply u-substitution

: \int \frac{u + 7}{u ^{10}} d u

= 72 \cdot \int \frac{u + 7}{u ^{10}} d u

Expand

\frac{u + 7}{u ^{10}} : \frac{1}{u ^{9}} + \frac{7}{u ^{10}}

= 72 \cdot \int \frac{1}{u ^{9}} + \frac{7}{u ^{10}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 72 (\int \frac{1}{u ^{9}} d u + \int \frac{7}{u ^{10}} d u)

\int \frac{1}{u ^{9}} d u = - \frac{1}{8 u ^{8}}

\int \frac{7}{u ^{10}} d u = - \frac{7}{9 u ^{9}}

= 72 (- \frac{1}{8 u ^{8}} - \frac{7}{9 u ^{9}})

Substitute back

u = x - 7

= 72 (- \frac{1}{8 ( x - 7 ) ^{8}} - \frac{7}{9 ( x - 7 ) ^{9}})

Add a constant to the solution

= 72 (- \frac{1}{8 ( x - 7 ) ^{8}} - \frac{7}{9 ( x - 7 ) ^{9}}) + C

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What is the integral of (72x)/((x-7)^{10)} ?
The integral of (72x)/((x-7)^{10)} is 72(-1/(8(x-7)^8)-7/(9(x-7)^9))+C