What is the integral of (8t^3-6t^{-2}) ?

The integral of (8t^3-6t^{-2}) is 2t^4+6/t+C

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integral of (8t^3-6t^{-2})

\int (8 t^{3} - 6 t^{- 2}) d t

2 t^{4} + \frac{6}{t} + C

Solution steps

\int 8 t^{3} - 6 t^{- 2} d t

6 t^{- 2} = \frac{6}{t ^{2}}

= \int 8 t^{3} - \frac{6}{t ^{2}} d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 8 t^{3} d t - \int \frac{6}{t ^{2}} d t

\int 8 t^{3} d t = 2 t^{4}

\int \frac{6}{t ^{2}} d t = - \frac{6}{t}

= 2 t^{4} - (- \frac{6}{t})

Simplify

= 2 t^{4} + \frac{6}{t}

Add a constant to the solution

= 2 t^{4} + \frac{6}{t} + C

x \to - 1^{- 1} lim (\frac{x}{x + 1})

\int \frac{9}{x + 9 + x} d x

d er i v a t i v e f (x) = x^{2} (1 - 8 x)

t an g e n t ln (2 x + 1)

f^{^{''}} (x) = \frac{x ^{2} - 2 x + 1}{x ^{2}}

What is the integral of (8t^3-6t^{-2}) ?
The integral of (8t^3-6t^{-2}) is 2t^4+6/t+C