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

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

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

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

2 ln x^{2} - 1 + C

Solution steps

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

Take the constant out:

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

= 4 \cdot \int \frac{x}{x ^{2} - 1} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

= 4 \cdot \int \frac{1}{2 u} d u

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} - 1

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

Simplify

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

= 2 ln x^{2} - 1

Add a constant to the solution

= 2 ln x^{2} - 1 + C

y^{^{''}} + 4 y^{^{'}} = 2 x - 3 e^{x}

\int 5 cos^{3} (x) d x

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

\frac{d}{d x} (\frac{1 + x}{1 - x})

(t^{3} - 3 t + 2) e^{- 2 t}

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