What is the integral of x/((x^2+6)4) ?

The integral of x/((x^2+6)4) is 1/8 ln|x^2+6|+C

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integral of x/((x^2+6)4)

integral

\frac{x}{( x ^{2} + 6 ) 4}

\frac{1}{8} ln x^{2} + 6 + C

Solution steps

\int \frac{x}{( x ^{2} + 6 ) \cdot 4} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{x}{x ^{2} + 6} d x

Apply u-substitution

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

= \frac{1}{4} \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

= \frac{1}{4} \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} + 6

= \frac{1}{4} \cdot \frac{1}{2} ln x^{2} + 6

Simplify

\frac{1}{4} \cdot \frac{1}{2} ln x^{2} + 6 : \frac{1}{8} ln x^{2} + 6

= \frac{1}{8} ln x^{2} + 6

Add a constant to the solution

= \frac{1}{8} ln x^{2} + 6 + C

x (lo g_{10} (x))^{2}

\frac{1}{x ^{2} + 1}

\frac{( x + 1 )}{( x + 2 ) ^{2}}

x^{3} (e^{- x})^{2}

\frac{1}{x ( k + x )}

What is the integral of x/((x^2+6)4) ?
The integral of x/((x^2+6)4) is 1/8 ln|x^2+6|+C