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

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

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

integral

e^{1 + x + l n (x)}

e^{1 + x} x - e^{1 + x} + C

Solution steps

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

Simplify

e^{1 + x + l n (x)} : e^{1 + x} x

= \int e^{1 + x} x d x

Apply Integration By Parts

: e^{1 + x} x - \int e^{1 + x} d x

= e^{1 + x} x - \int e^{1 + x} d x

\int e^{1 + x} d x = e^{1 + x}

= e^{1 + x} x - e^{1 + x}

Add a constant to the solution

= e^{1 + x} x - e^{1 + x} + C

\frac{x ^{2} - 4}{x}

\frac{x ^{2}}{4}

(2 x + 1)^{3}

2 + \frac{1}{x ^{2}} - \frac{3}{x ^{4}}

x^{2} + x \cdot x

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