What is the integral from 3 to 7 of 1/((7+2x)^3) ?

The integral from 3 to 7 of 1/((7+2x)^3) is 68/74529

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integral from 3 to 7 of 1/((7+2x)^3)

\int_{3}^{7} \frac{1}{( 7 + 2 x ) ^{3}} d x

\frac{68}{74529}

Decimal Notation

0.00091 \dots

Solution steps

\int_{3}^{7} \frac{1}{( 7 + 2 x ) ^{3}} d x

Apply u-substitution

: \int_{13}^{21} \frac{1}{2 u ^{3}} d u

= \int_{13}^{21} \frac{1}{2 u ^{3}} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{13}^{21} \frac{1}{u ^{3}} d u

Apply the Power Rule

: [- \frac{1}{2 u ^{2}}]_{13}^{21}

= \frac{1}{2} [- \frac{1}{2 u ^{2}}]_{13}^{21}

Compute the boundaries:

\frac{136}{74529}

= \frac{1}{2} \cdot \frac{136}{74529}

Simplify

= \frac{68}{74529}

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What is the integral from 3 to 7 of 1/((7+2x)^3) ?
The integral from 3 to 7 of 1/((7+2x)^3) is 68/74529