What is the integral of (x^2+x+1)/x ?

The integral of (x^2+x+1)/x is (x^2)/2+x+ln|x|+C

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integral of (x^2+x+1)/x

\int \frac{x ^{2} + x + 1}{x} d x

\frac{x ^{2}}{2} + x + ln ∣ x ∣ + C

Solution steps

\int \frac{x ^{2} + x + 1}{x} d x

Expand

\frac{x ^{2} + x + 1}{x} : x + 1 + \frac{1}{x}

= \int x + 1 + \frac{1}{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

= \int x d x + \int 1 d x + \int \frac{1}{x} d x

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

\int 1 d x = x

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{x ^{2}}{2} + x + ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{2}}{2} + x + ln ∣ x ∣ + C

(1 + x y) y^{^{'}} + y^{2} = 0

y = (arctan (5 x))^{2}

k = 0 \sum \infty (\frac{x - 4}{5})^{k}

\frac{\partial}{\partial x} (2 x sin (y - z))

\frac{3 e ^{(- 2 s)}}{( s ^{2} + 2 s + 5 )}

What is the integral of (x^2+x+1)/x ?
The integral of (x^2+x+1)/x is (x^2)/2+x+ln|x|+C