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

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

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

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

- \frac{1}{2} ln 4 + x^{2} + C

Solution steps

\int \frac{- x}{4 + 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}{4 + 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 = 4 + x^{2}

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

Add a constant to the solution

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

x \to 5 lim (- x^{2} + 9 x - 7)

\frac{d}{d x} ((3 x^{5} - 1)^{7})

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

\int_{1}^{2} (1 - t^{2}) t d t

f (x) = 4 x^{3} + 8 x, (- 1, - 12)

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