Solved: "Evaluate the integrals in ∫ln⁡(x)dxxln2(x)+1"

actever6a

Answered question

2021-11-10

Evaluate the integrals in

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

Pulad1971

Beginner2021-11-11Added 22 answers

Step 1
Consider the following integral:

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

Substitute

\ln^{2} (x) + 1 = u \to \frac{2 \ln (x)}{x} d x = d u

in the above integral:

I = \int \frac{1}{2 \sqrt{u}} d u

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

= \frac{1}{2} \times \frac{u^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} + C

= \frac{1}{2} \times \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C

= \sqrt{u} + C

Step 2
Substitute

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

in the above equation:

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

Hence, the solution is

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

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