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

The integral of (4x^5-16x^4+8x^3)/(4x^2) is 1/4 x^4-(4x^3)/3+x^2+C

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

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

\frac{1}{4} x^{4} - \frac{4 x ^{3}}{3} + x^{2} + C

Solution steps

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

Expand

\frac{4 x ^{5} - 16 x ^{4} + 8 x ^{3}}{x ^{2}} : \frac{4 x ^{5}}{x ^{2}} - \frac{16 x ^{4}}{x ^{2}} + \frac{8 x ^{3}}{x ^{2}}

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

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

\int \frac{16 x ^{4}}{x ^{2}} d x = \frac{16 x ^{3}}{3}

\int \frac{8 x ^{3}}{x ^{2}} d x = 4 x^{2}

= \frac{1}{4} (x^{4} - \frac{16 x ^{3}}{3} + 4 x^{2})

Simplify

\frac{1}{4} (x^{4} - \frac{16 x ^{3}}{3} + 4 x^{2}) : \frac{1}{4} x^{4} - \frac{4 x ^{3}}{3} + x^{2}

= \frac{1}{4} x^{4} - \frac{4 x ^{3}}{3} + x^{2}

Add a constant to the solution

= \frac{1}{4} x^{4} - \frac{4 x ^{3}}{3} + x^{2} + C

f (x) = \frac{4}{5} x^{10}

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

2 x^{3} + 6 x^{2} - 90 x - 6

\int 4 sin (x) d x

x \cdot 1 - x^{2}

What is the integral of (4x^5-16x^4+8x^3)/(4x^2) ?
The integral of (4x^5-16x^4+8x^3)/(4x^2) is 1/4 x^4-(4x^3)/3+x^2+C