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

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

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

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

- ln x^{2} + 1 + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

= 2 \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 (- \frac{1}{2} ln ∣ u ∣)

Substitute back

u = - x^{2} - 1

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

Simplify

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

= - ln x^{2} + 1

Add a constant to the solution

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

\int \frac{( 2 x ^{4} + 2 5 x )}{x} d x

t \to 0 lim (\frac{( 2 + t ) ^{3} - 8}{t})

x \to - 2 lim (4 x + 1)

\frac{d y}{d t} = - 5 (y - 2)

(x + 1)^{5}

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