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

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

\int \frac{( 3 t - 2 )}{t + 1} d t

\frac{3}{x ^{5}}

y^{^{'}} + 5 y = 15

\frac{\partial}{\partial x} (- x - sin (y^{2}))

\int x (cos (x) - sin (x)) d x

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