What is the integral of 1/(x(36+ln^2(x))) ?

The integral of 1/(x(36+ln^2(x))) is 1/6 arctan((ln(x))/6)+C

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integral of 1/(x(36+ln^2(x)))

\int \frac{1}{x ( 36 + ln ^{2} ( x ) )} d x

\frac{1}{6} arctan (\frac{ln ( x )}{6}) + C

Solution steps

\int \frac{1}{x ( 36 + ln ^{2} ( x ) )} d x

Apply u-substitution

: \int \frac{1}{36 + u ^{2}} d u

= \int \frac{1}{36 + u ^{2}} d u

Apply Integral Substitution

= \int \frac{1}{6 ( v ^{2} + 1 )} d v

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{6} arctan (v)

Substitute back

= \frac{1}{6} arctan (\frac{ln ( x )}{6})

Add a constant to the solution

= \frac{1}{6} arctan (\frac{ln ( x )}{6}) + C

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What is the integral of 1/(x(36+ln^2(x))) ?
The integral of 1/(x(36+ln^2(x))) is 1/6 arctan((ln(x))/6)+C