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

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

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

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

ln ln^{2} (x) + 1 + ln (x) + C

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

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

= ln u^{2} + 1 + u

Substitute back

u = ln (x)

= ln ln^{2} (x) + 1 + ln (x)

Add a constant to the solution

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

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