What is the integral from-7 to-4 of 2/((x+4)^3) ?

The integral from-7 to-4 of 2/((x+4)^3) is diverges

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integral from-7 to-4 of 2/((x+4)^3)

\int_{- 7}^{- 4} \frac{2}{( x + 4 ) ^{3}} d x

diverges

Solution steps

\int_{- 7}^{- 4} \frac{2}{( x + 4 ) ^{3}} d x

Take the constant out:

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

= 2 \cdot \int_{- 7}^{- 4} \frac{1}{( x + 4 ) ^{3}} d x

Apply u-substitution

: \int_{- 3}^{0} \frac{1}{u ^{3}} d u

= 2 \cdot \int_{- 3}^{0} \frac{1}{u ^{3}} d u

Apply the Power Rule

: [- \frac{1}{2 u ^{2}}]_{- 3}^{0}

= 2 [- \frac{1}{2 u ^{2}}]_{- 3}^{0}

Compute the boundaries:

diverges

= diverges

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What is the integral from-7 to-4 of 2/((x+4)^3) ?
The integral from-7 to-4 of 2/((x+4)^3) is diverges