What is the integral from 1 to e of (9x^2+9x+9)/x ?

The integral from 1 to e of (9x^2+9x+9)/x is 9e+(9(e^2-1))/2

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integral from 1 to e of (9x^2+9x+9)/x

\int_{1}^{e} \frac{9 x ^{2} + 9 x + 9}{x} d x

9 e + \frac{9 ( e ^{2} - 1 )}{2}

Decimal

53.21528 \dots

Solution steps

\int_{1}^{e} \frac{9 x ^{2} + 9 x + 9}{x} d x

Expand

\frac{9 x ^{2} + 9 x + 9}{x} : 9 x + 9 + \frac{9}{x}

= \int_{1}^{e} 9 x + 9 + \frac{9}{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_{1}^{e} 9 x d x + \int_{1}^{e} 9 d x + \int_{1}^{e} \frac{9}{x} d x

\int_{1}^{e} 9 x d x = \frac{9 ( e ^{2} - 1 )}{2}

\int_{1}^{e} 9 d x = 9 e - 9

\int_{1}^{e} \frac{9}{x} d x = 9

= \frac{9 ( e ^{2} - 1 )}{2} + 9 e - 9 + 9

Simplify

= 9 e + \frac{9 ( e ^{2} - 1 )}{2}

\int \frac{1}{6} d x

\frac{\partial}{\partial x} (6 x^{2} - 4 x y^{2})

f (- 1) = \frac{x + 3}{2}

\frac{\partial}{\partial x} (ln (1 - x - y) - x^{- 1} - y^{- 1})

f (x) = e^{- 5 x}

What is the integral from 1 to e of (9x^2+9x+9)/x ?
The integral from 1 to e of (9x^2+9x+9)/x is 9e+(9(e^2-1))/2