What is the integral of (sqrt(4-x^2))/x ?

The integral of (sqrt(4-x^2))/x is -ln| 1/2 sqrt(4-x^2)+1|+ln| 1/2 sqrt(4-x^2)-1|+sqrt(4-x^2)+C

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integral of (sqrt(4-x^2))/x

\int \frac{4 - x ^{2}}{x} d x

- ln \frac{1}{2} 4 - x^{2} + 1 + ln \frac{1}{2} 4 - x^{2} - 1 + 4 - x^{2} + C

Solution steps

\int \frac{4 - x ^{2}}{x} d x

Apply u-substitution

: \int - \frac{u ^{2}}{- u ^{2} + 4} d u

= \int - \frac{u ^{2}}{- u ^{2} + 4} d u

Take the constant out:

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

= - \int \frac{u ^{2}}{- u ^{2} + 4} d u

Apply Integral Substitution

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

Take the constant out:

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

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

\frac{v ^{2}}{- v ^{2} + 1} = \frac{1}{- v ^{2} + 1} - 1

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

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

\int 1 d v = v

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

Substitute back

= - 2 \frac{ln \frac{4 - x ^{2}}{2} + 1}{2} - \frac{ln \frac{4 - x ^{2}}{2} - 1}{2} - \frac{4 - x ^{2}}{2}

Simplify

- 2 \frac{ln \frac{4 - x ^{2}}{2} + 1}{2} - \frac{ln \frac{4 - x ^{2}}{2} - 1}{2} - \frac{4 - x ^{2}}{2} : - ln \frac{1}{2} 4 - x^{2} + 1 + ln \frac{1}{2} 4 - x^{2} - 1 + 4 - x^{2}

= - ln \frac{1}{2} 4 - x^{2} + 1 + ln \frac{1}{2} 4 - x^{2} - 1 + 4 - x^{2}

Add a constant to the solution

= - ln \frac{1}{2} 4 - x^{2} + 1 + ln \frac{1}{2} 4 - x^{2} - 1 + 4 - x^{2} + C

x \to - 2 lim (\frac{x}{( x + 2 ) ^{2}})

x \to - 4 lim ((x)^{2})

\int \frac{1}{x ^{2} + 3} d x

θ \to 0 lim (θ cos (\frac{3}{θ}))

\int \frac{( x - 3 ) ^{3}}{x} d x

What is the integral of (sqrt(4-x^2))/x ?
The integral of (sqrt(4-x^2))/x is -ln| 1/2 sqrt(4-x^2)+1|+ln| 1/2 sqrt(4-x^2)-1|+sqrt(4-x^2)+C