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

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

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

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

- \frac{1}{2} ln x^{2} + 1 + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= - \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} + 1

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

Add a constant to the solution

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

\frac{d}{d x} (2 x^{2} - 5 x + 3)

f (x) = 2 tan^{2} (6 x)

C (t) = 0.0225 t e^{- 0.0467 t}

\frac{d}{d x} (\frac{- 11}{( x - 3 ) ^{2}})

\frac{d}{d x} (x^{2} \cdot sin (x^{3}))

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