What is the integral of 1/(xsqrt((ln(2x))^2+9)) ?

The integral of 1/(xsqrt((ln(2x))^2+9)) is ln(1/3 |ln(2x)+sqrt(9+ln^2(2x))|)+C

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integral of 1/(xsqrt((ln(2x))^2+9))

\int \frac{1}{x ( ln ( 2 x ) ) ^{2} + 9} d x

ln (\frac{1}{3} ln (2 x) + 9 + ln^{2} (2 x)) + C

Solution steps

\int \frac{1}{x ( ln ( 2 x ) ) ^{2} + 9} d x

Apply u-substitution

: \int \frac{1}{u ln ^{2} ( u ) + 9} d u

= \int \frac{1}{u ln ^{2} ( u ) + 9} d u

Apply u-substitution

: \int \frac{1}{v ^{2} + 9} d v

= \int \frac{1}{v ^{2} + 9} d v

Apply Trigonometric Substitution

: \int sec (w) d w

= \int sec (w) d w

Use the common integral:

\int sec (w) d w = ln ∣ tan (w) + sec (w) ∣

= ln ∣ tan (w) + sec (w) ∣

Substitute back

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

Simplify

ln tan (arctan (\frac{1}{3} ln (2 x))) + sec (arctan (\frac{1}{3} ln (2 x))) : ln (\frac{1}{3} ln (2 x) + 9 + ln^{2} (2 x))

= ln (\frac{1}{3} ln (2 x) + 9 + ln^{2} (2 x))

Add a constant to the solution

= ln (\frac{1}{3} ln (2 x) + 9 + ln^{2} (2 x)) + C

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What is the integral of 1/(xsqrt((ln(2x))^2+9)) ?
The integral of 1/(xsqrt((ln(2x))^2+9)) is ln(1/3 |ln(2x)+sqrt(9+ln^2(2x))|)+C