What is the integral of [(4x-9)^6-2x] ?

The integral of [(4x-9)^6-2x] is 1/28 (4x-9)^7-x^2+C

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integral of [(4x-9)^6-2x]

\int [(4 x - 9)^{6} - 2 x] d x

\frac{1}{28} (4 x - 9)^{7} - x^{2} + C

Solution steps

\int (4 x - 9)^{6} - 2 x d x

Apply the Sum Rule:

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

= \int (4 x - 9)^{6} d x - \int 2 x d x

\int (4 x - 9)^{6} d x = \frac{1}{28} (4 x - 9)^{7}

\int 2 x d x = x^{2}

= \frac{1}{28} (4 x - 9)^{7} - x^{2}

Add a constant to the solution

= \frac{1}{28} (4 x - 9)^{7} - x^{2} + C

What is the integral of [(4x-9)^6-2x] ?
The integral of [(4x-9)^6-2x] is 1/28 (4x-9)^7-x^2+C