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

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

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

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

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

Decimal

0.34657 \dots

Solution steps

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

Apply u-substitution

= \int_{e^{2}}^{e^{4}} \frac{1}{2 u ln ( 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_{e^{2}}^{e^{4}} \frac{1}{u ln ( u )} d u

Apply u-substitution

= \frac{1}{2} \cdot \int_{2}^{4} \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} [ln ∣ v ∣]_{2}^{4}

Compute the boundaries:

ln (2)

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

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