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

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

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

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

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

Solution steps

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

Simplify

e^{x^{2} + l n (x)} : e^{x^{2}} x

= \int e^{x^{2}} x d x

Apply u-substitution

: \int \frac{e ^{u}}{2} d u

= \int \frac{e ^{u}}{2} 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^{u} d u

Use the common integral:

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

= \frac{1}{2} e^{u}

Substitute back

u = x^{2}

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

Add a constant to the solution

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

\frac{2}{x - 8}

\frac{d y}{d x} = \frac{5 x sec ( \frac{y}{x} ) + y}{x}

10 x - 6 y x^{2} + 1 y^{^{'}} = 0

x \to 0 lim (\frac{3 - cos ( x )}{x})

y^{^{''}} - 18 y^{^{'}} + 81 y = t^{2} e^{9 t} - e^{9 t}

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