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

The integral of (6+x^4)(7-x^3) is 42x-(3x^4)/2+(7x^5)/5-(x^8)/8+C

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integral of (6+x^4)(7-x^3)

\int (6 + x^{4}) (7 - x^{3}) d x

42 x - \frac{3 x ^{4}}{2} + \frac{7 x ^{5}}{5} - \frac{x ^{8}}{8} + C

Solution steps

\int (6 + x^{4}) (7 - x^{3}) d x

Expand

(6 + x^{4}) (7 - x^{3}) : 42 - 6 x^{3} + 7 x^{4} - x^{7}

= \int 42 - 6 x^{3} + 7 x^{4} - 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 42 d x - \int 6 x^{3} d x + \int 7 x^{4} d x - \int x^{7} d x

\int 42 d x = 42 x

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

\int 7 x^{4} d x = \frac{7 x ^{5}}{5}

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

= 42 x - \frac{3 x ^{4}}{2} + \frac{7 x ^{5}}{5} - \frac{x ^{8}}{8}

Add a constant to the solution

= 42 x - \frac{3 x ^{4}}{2} + \frac{7 x ^{5}}{5} - \frac{x ^{8}}{8} + C

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What is the integral of (6+x^4)(7-x^3) ?
The integral of (6+x^4)(7-x^3) is 42x-(3x^4)/2+(7x^5)/5-(x^8)/8+C