What is the integral from 1 to e^4 of 1/(x(1+ln(x))) ?

The integral from 1 to e^4 of 1/(x(1+ln(x))) is ln(5)

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integral from 1 to e^4 of 1/(x(1+ln(x)))

\int_{1}^{e^{4}} \frac{1}{x ( 1 + ln ( x ) )} d x

ln (5)

Decimal

1.60943 \dots

Solution steps

\int_{1}^{e^{4}} \frac{1}{x ( 1 + ln ( x ) )} d x

Apply u-substitution

= \int_{1}^{5} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{5}

Compute the boundaries:

ln (5)

= ln (5)

\int y^{2} \cdot cos (y) d y

\int \frac{5}{2 x - 2} d x

x \to 1 + lim (\frac{x}{ln ( x )})

x \to 3 lim (x^{2} + x - 5)

y^{^{''}} - (\frac{x}{x - 1}) y^{^{'}} + \frac{1}{x - 1} y = 0

What is the integral from 1 to e^4 of 1/(x(1+ln(x))) ?
The integral from 1 to e^4 of 1/(x(1+ln(x))) is ln(5)