What is the integral of 1/(4-(x+1)^2) ?

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

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integral of 1/(4-(x+1)^2)

\int \frac{1}{4 - ( x + 1 ) ^{2}} d x

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

Solution steps

\int \frac{1}{4 - ( x + 1 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{4 - u ^{2}} d u

= \int \frac{1}{4 - u ^{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}{4} (ln \frac{x + 1}{2} + 1 - ln \frac{x + 1}{2} - 1)

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

Add a constant to the solution

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

12 (t + 1) \frac{d y}{d t} - 9 y = 27 t

cos (4) x^{3}

\frac{d}{d x} (4 x y^{2} - x y^{3} - x^{2} y^{2})

cot^{2} (π x)

8 (2 z e^{z} + e^{z} z^{2})

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