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

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

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

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

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

Solution steps

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

Expand

2 - (x - 1)^{2} : - x^{2} + 2 x + 1

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

Complete the square

- x^{2} + 2 x + 1 : - (x - 1)^{2} + 2

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

Apply u-substitution

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

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

Apply Integral Substitution

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

Take the constant out:

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

= \frac{1}{2} \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}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d x} = k y (1 - y)

\int \frac{x ^{2} - 22}{x ^{3} - 2 x ^{2} + x} d x

x 8^{x}

\int 9 e^{- 9 x} sin (9 x) d x

x \to 1 lim (\frac{x ^{3}}{x ^{2} + 1})

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