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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{1}{v ^{2} - 1} d v

= \int \frac{1}{v ^{2} - 1} d v

Factor

v^{2} - 1 : - (- v^{2} + 1)

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= - (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

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

Simplify

- \frac{ln 1 + ln ^{2} ( x ) + 1}{2} - \frac{ln 1 + ln ^{2} ( x ) - 1}{2} : - \frac{1}{2} ln 1 + ln^{2} (x) + 1 + \frac{1}{2} ln 1 + ln^{2} (x) - 1

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

Add a constant to the solution

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

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