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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Factor

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

= 3 \cdot \int \frac{1}{- ( - u ^{2} + 1 )} d u

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

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

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

Add a constant to the solution

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

y = (x - 2)^{2} - 4, y = x + 1

\int \frac{cos ( \frac{π}{x} )}{x ^{2}} d x

\frac{d}{d x} (x^{2} - 16 x)

- 7 e^{(x + 2)} + e^{5}

(6, - 7), (8, 5)

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