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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

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 t^{- 3}

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

Simplify

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

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

Add a constant to the solution

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

\int t^{3} t^{4} + 5 d t

x \to 0 lim (x^{2} + 3 x - 1)

\frac{d}{d x} (4 x^{2} - 5)

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

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