What is the integral of (7y^6+8y^3-1) ?

The integral of (7y^6+8y^3-1) is y^7+2y^4-y+C

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integral of (7y^6+8y^3-1)

\int (7 y^{6} + 8 y^{3} - 1) d y

y^{7} + 2 y^{4} - y + C

Solution steps

\int 7 y^{6} + 8 y^{3} - 1 d y

Apply the Sum Rule:

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

= \int 7 y^{6} d y + \int 8 y^{3} d y - \int 1 d y

\int 7 y^{6} d y = y^{7}

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

\int 1 d y = y

= y^{7} + 2 y^{4} - y

Add a constant to the solution

= y^{7} + 2 y^{4} - y + C

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What is the integral of (7y^6+8y^3-1) ?
The integral of (7y^6+8y^3-1) is y^7+2y^4-y+C