What is the integral of (3x^{-4})(2+4x^{-3})^{-7} ?

The integral of (3x^{-4})(2+4x^{-3})^{-7} is (x^{18})/(24(2x^3+4)^6)+C

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integral of (3x^{-4})(2+4x^{-3})^{-7}

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

\frac{x ^{18}}{24 ( 2 x ^{3} + 4 ) ^{6}} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int - \frac{1}{12 u ^{7}} d u

= 3 \cdot \int - \frac{1}{12 u ^{7}} d u

Take the constant out:

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

= 3 (- \frac{1}{12} \cdot \int \frac{1}{u ^{7}} d u)

Apply the Power Rule

: - \frac{1}{6 u ^{6}}

= 3 (- \frac{1}{12} (- \frac{1}{6 u ^{6}}))

Substitute back

u = 2 + 4 x^{- 3}

= 3 (- \frac{1}{12} (- \frac{1}{6 ( 2 + 4 x ^{- 3} ) ^{6}}))

Simplify

3 (- \frac{1}{12} (- \frac{1}{6 ( 2 + 4 x ^{- 3} ) ^{6}})) : \frac{x ^{18}}{24 ( 2 x ^{3} + 4 ) ^{6}}

= \frac{x ^{18}}{24 ( 2 x ^{3} + 4 ) ^{6}}

Add a constant to the solution

= \frac{x ^{18}}{24 ( 2 x ^{3} + 4 ) ^{6}} + C

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What is the integral of (3x^{-4})(2+4x^{-3})^{-7} ?
The integral of (3x^{-4})(2+4x^{-3})^{-7} is (x^{18})/(24(2x^3+4)^6)+C