What is the integral of (ln(x)+1)e^{xln(x)} ?

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

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integral of (ln(x)+1)e^{xln(x)}

\int (ln (x) + 1) e^{x l n (x)} d x

x^{x} + C

Solution steps

\int (ln (x) + 1) e^{x l n (x)} d x

Apply u-substitution

= \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= e^{u}

Substitute back

u = x ln (x)

= e^{x l n (x)}

Simplify

e^{x l n (x)} : x^{x}

= x^{x}

Add a constant to the solution

= x^{x} + C

\frac{d}{d x} (ln (\frac{a ^{2} - x ^{2}}{x}))

x \to 3 lim (2 x^{2} + 1)

\frac{d}{d x} (3 x^{2} + 4 x + 2)

f (x) = \frac{x}{x ^{2} + 4}

d er i v a t i v e y = 3 + 9 - (x - 3)^{2}

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