What is the integral of 1/(4x+xln^2(3x)) ?

The integral of 1/(4x+xln^2(3x)) is 1/2 arctan((ln(3x))/2)+C

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integral of 1/(4x+xln^2(3x))

\int \frac{1}{4 x + x ln ^{2} ( 3 x )} d x

\frac{1}{2} arctan (\frac{ln ( 3 x )}{2}) + C

Solution steps

\int \frac{1}{4 x + x ln ^{2} ( 3 x )} d x

Apply u-substitution

= \int \frac{1}{4 u + u ln ^{2} ( u )} d u

Apply u-substitution

= \int \frac{1}{4 + v ^{2}} d v

Apply Integral Substitution

= \int \frac{1}{2 ( w ^{2} + 1 )} d w

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}{w ^{2} + 1} d w

Use the common integral:

\int \frac{1}{w ^{2} + 1} d w = arctan (w)

= \frac{1}{2} arctan (w)

Substitute back

= \frac{1}{2} arctan (\frac{ln ( 3 x )}{2})

Add a constant to the solution

= \frac{1}{2} arctan (\frac{ln ( 3 x )}{2}) + C

What is the integral of 1/(4x+xln^2(3x)) ?
The integral of 1/(4x+xln^2(3x)) is 1/2 arctan((ln(3x))/2)+C