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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

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

Add a constant to the solution

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

\int \frac{26}{26 + e ^{x}} d x

\frac{x}{x + 6}

\int_{6}^{8} \frac{24}{( x - 6 ) ^{3}} d x

f (x) = 5 x, (25, 25)

\frac{d ^{2} y}{d x ^{2}} + 6 \frac{d y}{d x} + 9 y = 0

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