What is the integral of 2/((x+1)^6) ?

The integral of 2/((x+1)^6) is -2/(5(x+1)^5)+C

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integral of 2/((x+1)^6)

\int \frac{2}{( x + 1 ) ^{6}} d x

- \frac{2}{5 ( x + 1 ) ^{5}} + C

Solution steps

\int \frac{2}{( x + 1 ) ^{6}} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{( x + 1 ) ^{6}} d x

Apply u-substitution

: \int \frac{1}{u ^{6}} d u

= 2 \cdot \int \frac{1}{u ^{6}} d u

Apply the Power Rule

: - \frac{1}{5 u ^{5}}

= 2 (- \frac{1}{5 u ^{5}})

Substitute back

u = x + 1

= 2 (- \frac{1}{5 ( x + 1 ) ^{5}})

Simplify

2 (- \frac{1}{5 ( x + 1 ) ^{5}}) : - \frac{2}{5 ( x + 1 ) ^{5}}

= - \frac{2}{5 ( x + 1 ) ^{5}}

Add a constant to the solution

= - \frac{2}{5 ( x + 1 ) ^{5}} + C

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What is the integral of 2/((x+1)^6) ?
The integral of 2/((x+1)^6) is -2/(5(x+1)^5)+C