What is the integral of (4x^4-2x^2+x-5)(4x^3-x+1/4) ?

The integral of (4x^4-2x^2+x-5)(4x^3-x+1/4) is 2x^8-2x^6+x^5-(9x^4)/2-(x^3)/2+(21x^2)/8-5/4 x+C

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integral of (4x^4-2x^2+x-5)(4x^3-x+1/4)

\int (4 x^{4} - 2 x^{2} + x - 5) (4 x^{3} - x + \frac{1}{4}) d x

2 x^{8} - 2 x^{6} + x^{5} - \frac{9 x ^{4}}{2} - \frac{x ^{3}}{2} + \frac{21 x ^{2}}{8} - \frac{5}{4} x + C

Solution steps

\int (4 x^{4} - 2 x^{2} + x - 5) (4 x^{3} - x + \frac{1}{4}) d x

Simplify

(4 x^{4} - 2 x^{2} + x - 5) (4 x^{3} - x + \frac{1}{4}) : 16 x^{7} - 12 x^{5} + 5 x^{4} - 18 x^{3} - \frac{3}{2} x^{2} + \frac{21}{4} x - \frac{5}{4}

= \int 16 x^{7} - 12 x^{5} + 5 x^{4} - 18 x^{3} - \frac{3}{2} x^{2} + \frac{21}{4} x - \frac{5}{4} 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 16 x^{7} d x - \int 12 x^{5} d x + \int 5 x^{4} d x - \int 18 x^{3} d x - \int \frac{3}{2} x^{2} d x + \int \frac{21}{4} x d x - \int \frac{5}{4} d x

\int 16 x^{7} d x = 2 x^{8}

\int 12 x^{5} d x = 2 x^{6}

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

\int 18 x^{3} d x = \frac{9 x ^{4}}{2}

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

\int \frac{21}{4} x d x = \frac{21 x ^{2}}{8}

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

= 2 x^{8} - 2 x^{6} + x^{5} - \frac{9 x ^{4}}{2} - \frac{x ^{3}}{2} + \frac{21 x ^{2}}{8} - \frac{5}{4} x

Add a constant to the solution

= 2 x^{8} - 2 x^{6} + x^{5} - \frac{9 x ^{4}}{2} - \frac{x ^{3}}{2} + \frac{21 x ^{2}}{8} - \frac{5}{4} x + C

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What is the integral of (4x^4-2x^2+x-5)(4x^3-x+1/4) ?
The integral of (4x^4-2x^2+x-5)(4x^3-x+1/4) is 2x^8-2x^6+x^5-(9x^4)/2-(x^3)/2+(21x^2)/8-5/4 x+C