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

= 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

= 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

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