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

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

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

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

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

Solution steps

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

y = \frac{8}{x - 7}

\frac{d}{d x} (2 x + cos (x))

\int x^{27} e^{- x^{28}} d x

\int_{- 2}^{5} ∣ x - 2 ∣ d x

(1 + \frac{10}{x})^{8}

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