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

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

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integral of (-3x^6+4x^5+4x^4+3)

\int (- 3 x^{6} + 4 x^{5} + 4 x^{4} + 3) d x

- \frac{3 x ^{7}}{7} + \frac{2 x ^{6}}{3} + \frac{4 x ^{5}}{5} + 3 x + C

Solution steps

\int - 3 x^{6} + 4 x^{5} + 4 x^{4} + 3 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^{5} d x + \int 4 x^{4} d x + \int 3 d x

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

\int 4 x^{5} d x = \frac{2 x ^{6}}{3}

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

\int 3 d x = 3 x

= - \frac{3 x ^{7}}{7} + \frac{2 x ^{6}}{3} + \frac{4 x ^{5}}{5} + 3 x

Add a constant to the solution

= - \frac{3 x ^{7}}{7} + \frac{2 x ^{6}}{3} + \frac{4 x ^{5}}{5} + 3 x + C

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