What is the integral of 1/2 t^{-3}(5t^{-2}+10)^{-4} ?

The integral of 1/2 t^{-3}(5t^{-2}+10)^{-4} is (t^6)/(60(5+10t^2)^3)+C

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integral of 1/2 t^{-3}(5t^{-2}+10)^{-4}

\int \frac{1}{2} t^{- 3} (5 t^{- 2} + 10)^{- 4} d t

\frac{t ^{6}}{60 ( 5 + 10 t ^{2} ) ^{3}} + C

Solution steps

\int \frac{1}{2} t^{- 3} (5 t^{- 2} + 10)^{- 4} d t

Take the constant out:

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

= \frac{1}{2} \cdot \int t^{- 3} (5 t^{- 2} + 10)^{- 4} d t

Apply u-substitution

: \int - \frac{1}{10 u ^{4}} d u

= \frac{1}{2} \cdot \int - \frac{1}{10 u ^{4}} d u

Take the constant out:

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

= \frac{1}{2} (- \frac{1}{10} \cdot \int \frac{1}{u ^{4}} d u)

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

= \frac{1}{2} (- \frac{1}{10} (- \frac{1}{3 u ^{3}}))

Substitute back

u = 5 t^{- 2} + 10

= \frac{1}{2} (- \frac{1}{10} (- \frac{1}{3 ( 5 t ^{- 2} + 10 ) ^{3}}))

Simplify

\frac{1}{2} (- \frac{1}{10} (- \frac{1}{3 ( 5 t ^{- 2} + 10 ) ^{3}})) : \frac{t ^{6}}{60 ( 5 + 10 t ^{2} ) ^{3}}

= \frac{t ^{6}}{60 ( 5 + 10 t ^{2} ) ^{3}}

Add a constant to the solution

= \frac{t ^{6}}{60 ( 5 + 10 t ^{2} ) ^{3}} + C

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What is the integral of 1/2 t^{-3}(5t^{-2}+10)^{-4} ?
The integral of 1/2 t^{-3}(5t^{-2}+10)^{-4} is (t^6)/(60(5+10t^2)^3)+C