What is the integral of x/((7x+1)^{17)} ?

The integral of x/((7x+1)^{17)} is (-112x-1)/(11760(7x+1)^{16)}+C

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integral of x/((7x+1)^{17)}

\int \frac{x}{( 7 x + 1 ) ^{17}} d x

\frac{- 112 x - 1}{11760 ( 7 x + 1 ) ^{16}} + C

Solution steps

\int \frac{x}{( 7 x + 1 ) ^{17}} d x

Apply u-substitution

: \int \frac{u - 1}{49 u ^{17}} d u

= \int \frac{u - 1}{49 u ^{17}} d u

Take the constant out:

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

= \frac{1}{49} \cdot \int \frac{u - 1}{u ^{17}} d u

Expand

\frac{u - 1}{u ^{17}} : \frac{1}{u ^{16}} - \frac{1}{u ^{17}}

= \frac{1}{49} \cdot \int \frac{1}{u ^{16}} - \frac{1}{u ^{17}} 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

= \frac{1}{49} (\int \frac{1}{u ^{16}} d u - \int \frac{1}{u ^{17}} d u)

\int \frac{1}{u ^{16}} d u = - \frac{1}{15 u ^{15}}

\int \frac{1}{u ^{17}} d u = - \frac{1}{16 u ^{16}}

= \frac{1}{49} (- \frac{1}{15 u ^{15}} - (- \frac{1}{16 u ^{16}}))

Substitute back

u = 7 x + 1

= \frac{1}{49} (- \frac{1}{15 ( 7 x + 1 ) ^{15}} - (- \frac{1}{16 ( 7 x + 1 ) ^{16}}))

Simplify

\frac{1}{49} (- \frac{1}{15 ( 7 x + 1 ) ^{15}} - (- \frac{1}{16 ( 7 x + 1 ) ^{16}})) : \frac{- 112 x - 1}{11760 ( 7 x + 1 ) ^{16}}

= \frac{- 112 x - 1}{11760 ( 7 x + 1 ) ^{16}}

Add a constant to the solution

= \frac{- 112 x - 1}{11760 ( 7 x + 1 ) ^{16}} + C

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What is the integral of x/((7x+1)^{17)} ?
The integral of x/((7x+1)^{17)} is (-112x-1)/(11760(7x+1)^{16)}+C