What is the integral of (4-x^3)(7+x^4) ?

The integral of (4-x^3)(7+x^4) is 28x+(4x^5)/5-(7x^4)/4-(x^8)/8+C

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integral of (4-x^3)(7+x^4)

\int (4 - x^{3}) (7 + x^{4}) d x

28 x + \frac{4 x ^{5}}{5} - \frac{7 x ^{4}}{4} - \frac{x ^{8}}{8} + C

Solution steps

\int (4 - x^{3}) (7 + x^{4}) d x

Expand

(4 - x^{3}) (7 + x^{4}) : 28 + 4 x^{4} - 7 x^{3} - x^{7}

= \int 28 + 4 x^{4} - 7 x^{3} - x^{7} 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 28 d x + \int 4 x^{4} d x - \int 7 x^{3} d x - \int x^{7} d x

\int 28 d x = 28 x

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

\int 7 x^{3} d x = \frac{7 x ^{4}}{4}

\int x^{7} d x = \frac{x ^{8}}{8}

= 28 x + \frac{4 x ^{5}}{5} - \frac{7 x ^{4}}{4} - \frac{x ^{8}}{8}

Add a constant to the solution

= 28 x + \frac{4 x ^{5}}{5} - \frac{7 x ^{4}}{4} - \frac{x ^{8}}{8} + C

What is the integral of (4-x^3)(7+x^4) ?
The integral of (4-x^3)(7+x^4) is 28x+(4x^5)/5-(7x^4)/4-(x^8)/8+C