What is the integral of (x+xln(x))^{-1} ?

The integral of (x+xln(x))^{-1} is ln|1+ln(x)|+C

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integral of (x+xln(x))^{-1}

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

ln ∣ 1 + ln (x) ∣ + C

Solution steps

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

Apply exponent rule:

a^{- 1} = \frac{1}{a}

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

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{1}{v} d v

= \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= ln ∣ v ∣

Substitute back

= ln ∣ 1 + ln (x) ∣

Add a constant to the solution

= ln ∣ 1 + ln (x) ∣ + C

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What is the integral of (x+xln(x))^{-1} ?
The integral of (x+xln(x))^{-1} is ln|1+ln(x)|+C