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

The integral of-4x^6+6x^4+3x^3+6 is -(4x^7)/7+(6x^5)/5+(3x^4)/4+6x+C

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integral of-4x^6+6x^4+3x^3+6

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

- \frac{4 x ^{7}}{7} + \frac{6 x ^{5}}{5} + \frac{3 x ^{4}}{4} + 6 x + C

Solution steps

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

\int 4 x^{6} d x = \frac{4 x ^{7}}{7}

\int 6 x^{4} d x = \frac{6 x ^{5}}{5}

\int 3 x^{3} d x = \frac{3 x ^{4}}{4}

\int 6 d x = 6 x

= - \frac{4 x ^{7}}{7} + \frac{6 x ^{5}}{5} + \frac{3 x ^{4}}{4} + 6 x

Add a constant to the solution

= - \frac{4 x ^{7}}{7} + \frac{6 x ^{5}}{5} + \frac{3 x ^{4}}{4} + 6 x + C

\int 6 x x - 5 d x

\int (5 x + 1) d x

\int 5 e^{- 5 t} d t

\int \frac{cos ^{3} ( t )}{sin ( t )} d t

\int x (3 x + 1) d x

What is the integral of-4x^6+6x^4+3x^3+6 ?
The integral of-4x^6+6x^4+3x^3+6 is -(4x^7)/7+(6x^5)/5+(3x^4)/4+6x+C