What is the integral of 1/(3-(x-2)^2) ?

The integral of 1/(3-(x-2)^2) is 1/(2sqrt(3))(ln|(x-2)/(sqrt(3))+1|-ln|(x-2)/(sqrt(3))-1|)+C

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integral of 1/(3-(x-2)^2)

\int \frac{1}{3 - ( x - 2 ) ^{2}} d x

\frac{1}{2 3} (ln \frac{x - 2}{3} + 1 - ln \frac{x - 2}{3} - 1) + C

Solution steps

\int \frac{1}{3 - ( x - 2 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{3 - u ^{2}} d u

= \int \frac{1}{3 - u ^{2}} d u

Apply Integral Substitution

= \int \frac{1}{3 ( 1 - v ^{2} )} d v

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{1 - v ^{2}} d v

Use the common integral:

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

= \frac{1}{3} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= \frac{1}{3} \frac{ln \frac{x - 2}{3} + 1}{2} - \frac{ln \frac{x - 2}{3} - 1}{2}

Simplify

\frac{1}{3} \frac{ln \frac{x - 2}{3} + 1}{2} - \frac{ln \frac{x - 2}{3} - 1}{2} : \frac{1}{2 3} (ln \frac{x - 2}{3} + 1 - ln \frac{x - 2}{3} - 1)

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

Add a constant to the solution

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

\int 4 - ∣ x ∣ d x

\int_{- 1}^{1} x + 1 - 1 d x

\int_{- 3}^{2} ∣ x + 1 ∣ d x

x \to 1 lim (\frac{2 x}{ln ( x )})

(2.9) (9.7)

What is the integral of 1/(3-(x-2)^2) ?
The integral of 1/(3-(x-2)^2) is 1/(2sqrt(3))(ln|(x-2)/(sqrt(3))+1|-ln|(x-2)/(sqrt(3))-1|)+C