What is the integral of (16x^3+12x^2+5x+8)/4 ?

The integral of (16x^3+12x^2+5x+8)/4 is x^4+x^3+(5x^2)/8+2x+C

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integral of (16x^3+12x^2+5x+8)/4

\int \frac{16 x ^{3} + 12 x ^{2} + 5 x + 8}{4} d x

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

Solution steps

\int \frac{16 x ^{3} + 12 x ^{2} + 5 x + 8}{4} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int 16 x^{3} + 12 x^{2} + 5 x + 8 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

= \frac{1}{4} (\int 16 x^{3} d x + \int 12 x^{2} d x + \int 5 x d x + \int 8 d x)

\int 16 x^{3} d x = 4 x^{4}

\int 12 x^{2} d x = 4 x^{3}

\int 5 x d x = \frac{5 x ^{2}}{2}

\int 8 d x = 8 x

= \frac{1}{4} (4 x^{4} + 4 x^{3} + \frac{5 x ^{2}}{2} + 8 x)

Simplify

\frac{1}{4} (4 x^{4} + 4 x^{3} + \frac{5 x ^{2}}{2} + 8 x) : x^{4} + x^{3} + \frac{5 x ^{2}}{8} + 2 x

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

Add a constant to the solution

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

\int 5^{5 x} d x

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What is the integral of (16x^3+12x^2+5x+8)/4 ?
The integral of (16x^3+12x^2+5x+8)/4 is x^4+x^3+(5x^2)/8+2x+C