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

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

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

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

\frac{1}{8} ln 1 + 4 ln^{2} (x) + C

Solution steps

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

Simplify

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

Apply u-substitution

: \int \frac{u}{1 + 4 u ^{2}} d u

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

Apply u-substitution

: \int \frac{1}{8 v} d v

= \int \frac{1}{8 v} d v

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{8} ln ∣ v ∣

Substitute back

= \frac{1}{8} ln 1 + 4 ln^{2} (x)

Add a constant to the solution

= \frac{1}{8} ln 1 + 4 ln^{2} (x) + C

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