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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \int \frac{1}{2 + u ^{2}} d u

Apply Integral Substitution

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

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}{1 + v ^{2}} d v

Use the common integral:

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

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

Substitute back

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

Add a constant to the solution

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

rr (θ) 2 - 4 sin (θ)

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