What is the integral of (ln(x))/x e^{ln^2(x)} ?

The integral of (ln(x))/x e^{ln^2(x)} is 1/2 e^{ln^2(x)}+C

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integral of (ln(x))/x e^{ln^2(x)}

\int \frac{ln ( x )}{x} e^{l n^{2} (x)} d x

\frac{1}{2} e^{l n^{2} (x)} + C

Solution steps

\int \frac{ln ( x )}{x} e^{l n^{2} (x)} d x

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \int \frac{e ^{l n^{2} (x)} ln ( x )}{x} d x

Apply u-substitution

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

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

Integral of a constant:

\int a d x = a x

= \frac{1}{2} u

Substitute back

u = e^{l n^{2} (x)}

= \frac{1}{2} e^{l n^{2} (x)}

Add a constant to the solution

= \frac{1}{2} e^{l n^{2} (x)} + C

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What is the integral of (ln(x))/x e^{ln^2(x)} ?
The integral of (ln(x))/x e^{ln^2(x)} is 1/2 e^{ln^2(x)}+C