What is the integral of 1/(x^2sqrt(25-x^2))x) ?

The integral of 1/(x^2sqrt(25-x^2))x) is -d/(10)(ln|(sqrt(-x^2+25))/5+1|-ln|(sqrt(-x^2+25))/5-1|)+C

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integral of 1/(x^2sqrt(25-x^2))x)

\int \frac{1}{x ^{2} 25 - x ^{2}} d (x) d x

- \frac{d}{10} (ln \frac{- x ^{2} + 25}{5} + 1 - ln \frac{- x ^{2} + 25}{5} - 1) + C

Solution steps

\int \frac{1}{x ^{2} 25 - x ^{2}} d x d x

Take the constant out:

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

= d \cdot \int \frac{1}{x ^{2} 25 - x ^{2}} x d x

Simplify

\frac{1}{x ^{2} 25 - x ^{2}} x : \frac{1}{x - x ^{2} + 25}

= d \cdot \int \frac{1}{x - x ^{2} + 25} d x

Apply u-substitution

: \int - \frac{1}{- u ^{2} + 25} d u

= d \cdot \int - \frac{1}{- u ^{2} + 25} d u

Take the constant out:

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

= d (- \int \frac{1}{- u ^{2} + 25} d u)

Apply Integral Substitution

= d (- \int \frac{1}{5 ( - v ^{2} + 1 )} d v)

Take the constant out:

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

= d (- \frac{1}{5} \cdot \int \frac{1}{- v ^{2} + 1} d v)

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= d (- \frac{1}{5} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= d - \frac{1}{5} \frac{ln \frac{- x ^{2} + 25}{5} + 1}{2} - \frac{ln \frac{- x ^{2} + 25}{5} - 1}{2}

Simplify

d - \frac{1}{5} \frac{ln \frac{- x ^{2} + 25}{5} + 1}{2} - \frac{ln \frac{- x ^{2} + 25}{5} - 1}{2} : - \frac{d}{10} (ln \frac{- x ^{2} + 25}{5} + 1 - ln \frac{- x ^{2} + 25}{5} - 1)

= - \frac{d}{10} (ln \frac{- x ^{2} + 25}{5} + 1 - ln \frac{- x ^{2} + 25}{5} - 1)

Add a constant to the solution

= - \frac{d}{10} (ln \frac{- x ^{2} + 25}{5} + 1 - ln \frac{- x ^{2} + 25}{5} - 1) + C

\int_{0}^{1} \frac{97}{x ^{5}} d x

\frac{d N}{d t} = 0.2 (4000 - N)

\int 2 x + 9 e^{x} d x

16 - x^{2}

\frac{\partial}{\partial x} (x^{2} + 16)

What is the integral of 1/(x^2sqrt(25-x^2))x) ?
The integral of 1/(x^2sqrt(25-x^2))x) is -d/(10)(ln|(sqrt(-x^2+25))/5+1|-ln|(sqrt(-x^2+25))/5-1|)+C