What is the integral of (3x^6-4x) ?

The integral of (3x^6-4x) is (3x^7)/7-2x^2+C

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integral of (3x^6-4x)

\int (3 x^{6} - 4 x) d x

\frac{3 x ^{7}}{7} - 2 x^{2} + C

Solution steps

\int 3 x^{6} - 4 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 3 x^{6} d x - \int 4 x d x

\int 3 x^{6} d x = \frac{3 x ^{7}}{7}

\int 4 x d x = 2 x^{2}

= \frac{3 x ^{7}}{7} - 2 x^{2}

Add a constant to the solution

= \frac{3 x ^{7}}{7} - 2 x^{2} + C

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What is the integral of (3x^6-4x) ?
The integral of (3x^6-4x) is (3x^7)/7-2x^2+C