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

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

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integral of x^{2x}(ln(x)+1)

\int x^{2 x} (ln (x) + 1) d x

\frac{1}{2} x^{2 x} + C

Solution steps

\int x^{2 x} (ln (x) + 1) 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 = x^{2 x}

= \frac{1}{2} x^{2 x}

Add a constant to the solution

= \frac{1}{2} x^{2 x} + C

\frac{d}{d t} (2 t^{2})

x \to - 8 - lim (x^{2} + \frac{9}{x + 8})

\frac{x ^{6}}{6} + \frac{1}{16 x ^{4}}

sin (x) y^{^{'}} + cos (x) y = 2

f (x) = \frac{( x ^{2} )}{( 5 + x )}

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