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

The integral from 1 to e of 1/(x(1+ln(x^2))) is 1/2 ln(3)

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integral from 1 to e of 1/(x(1+ln(x^2)))

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

\frac{1}{2} ln (3)

Decimal Notation

0.54930 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{3} \frac{1}{2 u} d u

= \int_{1}^{3} \frac{1}{2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{1}^{3} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{1}^{3}

Compute the boundaries:

ln (3)

= \frac{1}{2} ln (3)

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What is the integral from 1 to e of 1/(x(1+ln(x^2))) ?
The integral from 1 to e of 1/(x(1+ln(x^2))) is 1/2 ln(3)