What is the integral from 1 to 9 of (sqrt(u)-2u^2)/u ?

The integral from 1 to 9 of (sqrt(u)-2u^2)/u is -76

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integral from 1 to 9 of (sqrt(u)-2u^2)/u

\int_{1}^{9} \frac{u - 2 u ^{2}}{u} d u

- 76

Solution steps

\int_{1}^{9} \frac{u - 2 u ^{2}}{u} d u

Expand

\frac{u - 2 u ^{2}}{u} : \frac{1}{u} - 2 u

= \int_{1}^{9} \frac{1}{u} - 2 u d u

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}^{9} \frac{1}{u} d u - \int_{1}^{9} 2 u d u

\int_{1}^{9} \frac{1}{u} d u = 4

\int_{1}^{9} 2 u d u = 80

= 4 - 80

Simplify

= - 76

\int_{0}^{2} \frac{x}{x ^{2} - 1} d x

\int_{0}^{π} 1 + sin (2 x) d x

\int_{0}^{\frac{π}{2}} \frac{1}{1 - sin ( x )} d x

\int_{0}^{4} x - 1 d x

\int_{1}^{4} x^{\frac{1}{2}} d x

What is the integral from 1 to 9 of (sqrt(u)-2u^2)/u ?
The integral from 1 to 9 of (sqrt(u)-2u^2)/u is -76