What is the integral of 1/((4x+4)^3) ?

The integral of 1/((4x+4)^3) is x/((4x+4)^3)+(-3x-1)/(128(x+1)^3)+C

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integral of 1/((4x+4)^3)

\int \frac{1}{( 4 x + 4 ) ^{3}} d x

\frac{x}{( 4 x + 4 ) ^{3}} + \frac{- 3 x - 1}{128 ( x + 1 ) ^{3}} + C

Solution steps

\int \frac{1}{( 4 x + 4 ) ^{3}} d x

Apply Integration By Parts

: \frac{x}{( 4 x + 4 ) ^{3}} - \int - \frac{3 x}{64 ( x + 1 ) ^{4}} d x

= \frac{x}{( 4 x + 4 ) ^{3}} - \int - \frac{3 x}{64 ( x + 1 ) ^{4}} d x

\int - \frac{3 x}{64 ( x + 1 ) ^{4}} d x = - \frac{- 3 x - 1}{128 ( x + 1 ) ^{3}}

= \frac{x}{( 4 x + 4 ) ^{3}} - (- \frac{- 3 x - 1}{128 ( x + 1 ) ^{3}})

Simplify

= \frac{x}{( 4 x + 4 ) ^{3}} + \frac{- 3 x - 1}{128 ( x + 1 ) ^{3}}

Add a constant to the solution

= \frac{x}{( 4 x + 4 ) ^{3}} + \frac{- 3 x - 1}{128 ( x + 1 ) ^{3}} + C

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What is the integral of 1/((4x+4)^3) ?
The integral of 1/((4x+4)^3) is x/((4x+4)^3)+(-3x-1)/(128(x+1)^3)+C