What is the integral of (x^4+10x^3)/(2x^2) ?

The integral of (x^4+10x^3)/(2x^2) is 1/2 ((x^3)/3+5x^2)+C

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integral of (x^4+10x^3)/(2x^2)

\int \frac{x ^{4} + 10 x ^{3}}{2 x ^{2}} d x

\frac{1}{2} (\frac{x ^{3}}{3} + 5 x^{2}) + C

Solution steps

\int \frac{x ^{4} + 10 x ^{3}}{2 x ^{2}} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{x ^{4} + 10 x ^{3}}{x ^{2}} d x

\frac{x ^{4} + 10 x ^{3}}{x ^{2}} = x^{2} + 10 x

= \frac{1}{2} \cdot \int x^{2} + 10 x 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}{2} (\int x^{2} d x + \int 10 x d x)

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

\int 10 x d x = 5 x^{2}

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

Add a constant to the solution

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

\int 2 (1 - 3 x)^{2} d x

\frac{d}{d x} ((x^{3} - 2)^{2})

\int \frac{1}{e ^{- x} + e ^{x}} d x

t \to 0 + lim (\frac{4}{t} - \frac{1}{e ^{t} - 1})

\frac{d}{d x} (2 csc (x) cot (x))

What is the integral of (x^4+10x^3)/(2x^2) ?
The integral of (x^4+10x^3)/(2x^2) is 1/2 ((x^3)/3+5x^2)+C