What is the integral of (48)/((6x+1)^9) ?

The integral of (48)/((6x+1)^9) is (48x)/((6x+1)^9)+(-54x-1)/((6x+1)^9)+C

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integral of (48)/((6x+1)^9)

\int \frac{48}{( 6 x + 1 ) ^{9}} d x

\frac{48 x}{( 6 x + 1 ) ^{9}} + \frac{- 54 x - 1}{( 6 x + 1 ) ^{9}} + C

Solution steps

\int \frac{48}{( 6 x + 1 ) ^{9}} d x

Take the constant out:

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

= 48 \cdot \int \frac{1}{( 6 x + 1 ) ^{9}} d x

Apply Integration By Parts

: \frac{x}{( 6 x + 1 ) ^{9}} - \int - \frac{54 x}{( 6 x + 1 ) ^{10}} d x

= 48 (\frac{x}{( 6 x + 1 ) ^{9}} - \int - \frac{54 x}{( 6 x + 1 ) ^{10}} d x)

\int - \frac{54 x}{( 6 x + 1 ) ^{10}} d x = - \frac{- 54 x - 1}{48 ( 6 x + 1 ) ^{9}}

= 48 (\frac{x}{( 6 x + 1 ) ^{9}} - (- \frac{- 54 x - 1}{48 ( 6 x + 1 ) ^{9}}))

Simplify

48 (\frac{x}{( 6 x + 1 ) ^{9}} - (- \frac{- 54 x - 1}{48 ( 6 x + 1 ) ^{9}})) : \frac{48 x}{( 6 x + 1 ) ^{9}} + \frac{- 54 x - 1}{( 6 x + 1 ) ^{9}}

= \frac{48 x}{( 6 x + 1 ) ^{9}} + \frac{- 54 x - 1}{( 6 x + 1 ) ^{9}}

Add a constant to the solution

= \frac{48 x}{( 6 x + 1 ) ^{9}} + \frac{- 54 x - 1}{( 6 x + 1 ) ^{9}} + C

\int 10 sin^{3} (x co) s^{7} x d x

\frac{4 x}{x}

\int \frac{x}{2 x - 6} d x

4 x y^{^{'}} - 20 y = x^{- 8}

\frac{d}{d x} ((c x)^{\frac{1}{2}})

What is the integral of (48)/((6x+1)^9) ?
The integral of (48)/((6x+1)^9) is (48x)/((6x+1)^9)+(-54x-1)/((6x+1)^9)+C