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

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

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

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

ln ∣ ln (x) ∣ - ln ∣ ln (x) + 1 ∣ + C

Solution steps

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

Apply u-substitution

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

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

Take the partial fraction of

\frac{1}{u + u ^{2}} : \frac{1}{u} - \frac{1}{u + 1}

= \int \frac{1}{u} - \frac{1}{u + 1} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{u} d u - \int \frac{1}{u + 1} d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u + 1} d u = ln ∣ u + 1 ∣

= ln ∣ u ∣ - ln ∣ u + 1 ∣

Substitute back

u = ln (x)

= ln ∣ ln (x) ∣ - ln ∣ ln (x) + 1 ∣

Add a constant to the solution

= ln ∣ ln (x) ∣ - ln ∣ ln (x) + 1 ∣ + C

x \to 2 lim (7)

(2 x - 3)^{4} (x^{2} + x + 1)^{5}

\frac{d}{d t} (A t e^{t})

n = 1 \sum \infty \frac{n}{1 - 7 n}

\frac{d}{d x} (\frac{- 6 x}{2 x - 1})

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