What is the integral of (8t^4+5t^3-6t^2+7t-9) ?

The integral of (8t^4+5t^3-6t^2+7t-9) is (8t^5)/5+(5t^4)/4-2t^3+(7t^2)/2-9t+C

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integral of (8t^4+5t^3-6t^2+7t-9)

\int (8 t^{4} + 5 t^{3} - 6 t^{2} + 7 t - 9) d t

\frac{8 t ^{5}}{5} + \frac{5 t ^{4}}{4} - 2 t^{3} + \frac{7 t ^{2}}{2} - 9 t + C

Solution steps

\int 8 t^{4} + 5 t^{3} - 6 t^{2} + 7 t - 9 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^{4} d t + \int 5 t^{3} d t - \int 6 t^{2} d t + \int 7 t d t - \int 9 d t

\int 8 t^{4} d t = \frac{8 t ^{5}}{5}

\int 5 t^{3} d t = \frac{5 t ^{4}}{4}

\int 6 t^{2} d t = 2 t^{3}

\int 7 t d t = \frac{7 t ^{2}}{2}

\int 9 d t = 9 t

= \frac{8 t ^{5}}{5} + \frac{5 t ^{4}}{4} - 2 t^{3} + \frac{7 t ^{2}}{2} - 9 t

Add a constant to the solution

= \frac{8 t ^{5}}{5} + \frac{5 t ^{4}}{4} - 2 t^{3} + \frac{7 t ^{2}}{2} - 9 t + C

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\frac{d}{d x} (- e^{- x} sin (x) + e^{- x} cos (x))

\frac{\partial}{\partial x} (y z^{2} cos (x))

\int_{- 8}^{8} \frac{( 64 - x ^{2} )}{2} d x

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What is the integral of (8t^4+5t^3-6t^2+7t-9) ?
The integral of (8t^4+5t^3-6t^2+7t-9) is (8t^5)/5+(5t^4)/4-2t^3+(7t^2)/2-9t+C