What is the integral from 9 to 25 of (sqrt(x-9))/x ?

The integral from 9 to 25 of (sqrt(x-9))/x is 6(-arctan(4/3)+4/3)

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integral from 9 to 25 of (sqrt(x-9))/x

\int_{9}^{25} \frac{x - 9}{x} d x

6 (- arctan (\frac{4}{3}) + \frac{4}{3})

Solution steps

\int_{9}^{25} \frac{x - 9}{x} d x

Apply u-substitution

: \int_{0}^{4} \frac{2 u ^{2}}{u ^{2} + 9} d u

= \int_{0}^{4} \frac{2 u ^{2}}{u ^{2} + 9} d u

Take the constant out:

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

= 2 \cdot \int_{0}^{4} \frac{u ^{2}}{u ^{2} + 9} d u

\frac{u ^{2}}{u ^{2} + 9} = - \frac{9}{u ^{2} + 9} + 1

= 2 \cdot \int_{0}^{4} - \frac{9}{u ^{2} + 9} + 1 d u

Apply Integral Substitution

= 2 \cdot \int_{0}^{\frac{4}{3}} 3 (- \frac{1}{v ^{2} + 1} + 1) d v

Take the constant out:

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

= 2 \cdot 3 \cdot \int_{0}^{\frac{4}{3}} - \frac{1}{v ^{2} + 1} + 1 d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 \cdot 3 (- \int_{0}^{\frac{4}{3}} \frac{1}{v ^{2} + 1} d v + \int_{0}^{\frac{4}{3}} 1 d v)

\int_{0}^{\frac{4}{3}} \frac{1}{v ^{2} + 1} d v = arctan (\frac{4}{3})

\int_{0}^{\frac{4}{3}} 1 d v = \frac{4}{3}

= 2 \cdot 3 (- arctan (\frac{4}{3}) + \frac{4}{3})

Simplify

= 6 (- arctan (\frac{4}{3}) + \frac{4}{3})

(cos (x) + y sin (x)) d x + (cos (x)) d y = 0

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

y = 6 sec (x), at x = \frac{π}{6}

\int cos^{4} (d) x d x

x^{2} y^{^{''}} + 3 x y^{^{'}} + 3 y = 0

What is the integral from 9 to 25 of (sqrt(x-9))/x ?
The integral from 9 to 25 of (sqrt(x-9))/x is 6(-arctan(4/3)+4/3)