What is the integral of (3x)/(5+6x^2) ?

The integral of (3x)/(5+6x^2) is 1/4 ln|5+6x^2|+C

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integral of (3x)/(5+6x^2)

\int \frac{3 x}{5 + 6 x ^{2}} d x

\frac{1}{4} ln 5 + 6 x^{2} + C

Solution steps

\int \frac{3 x}{5 + 6 x ^{2}} d x

Take the constant out:

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

= 3 \cdot \int \frac{x}{5 + 6 x ^{2}} d x

Apply u-substitution

: \int \frac{1}{12 u} d u

= 3 \cdot \int \frac{1}{12 u} d u

Take the constant out:

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

= 3 \cdot \frac{1}{12} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{12} ln ∣ u ∣

Substitute back

u = 5 + 6 x^{2}

= 3 \cdot \frac{1}{12} ln 5 + 6 x^{2}

Simplify

3 \cdot \frac{1}{12} ln 5 + 6 x^{2} : \frac{1}{4} ln 5 + 6 x^{2}

= \frac{1}{4} ln 5 + 6 x^{2}

Add a constant to the solution

= \frac{1}{4} ln 5 + 6 x^{2} + C

\int \frac{e ^{x}}{1 + 4 e ^{2 x}} d x

d x - \frac{1}{y ^{2} - 6 y + 13} d y = 0

cos (2 x)

\frac{d}{d x} (sec (x) sin (x))

\int \frac{ln ( x )}{x ^{2}} d x

What is the integral of (3x)/(5+6x^2) ?
The integral of (3x)/(5+6x^2) is 1/4 ln|5+6x^2|+C