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

The integral of x^2ln(x)+e^{2x}+1 is 1/3 x^3ln(x)-(x^3)/9+1/2 e^{2x}+x+C

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

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

\frac{1}{3} x^{3} ln (x) - \frac{x ^{3}}{9} + \frac{1}{2} e^{2 x} + x + C

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int x^{2} ln (x) d x = \frac{1}{3} x^{3} ln (x) - \frac{x ^{3}}{9}

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

\int 1 d x = x

= \frac{1}{3} x^{3} ln (x) - \frac{x ^{3}}{9} + \frac{1}{2} e^{2 x} + x

Add a constant to the solution

= \frac{1}{3} x^{3} ln (x) - \frac{x ^{3}}{9} + \frac{1}{2} e^{2 x} + x + C

\frac{d}{d x} (x^{2} - \frac{1}{x})

e^{- 5 x}

\frac{d}{d x} (x^{2} y^{4} e^{y})

\frac{d}{d x} (\frac{x ^{2}}{x ^{2} - 6})

6 - x^{2}, x^{2} - 20 < x < 3

What is the integral of x^2ln(x)+e^{2x}+1 ?
The integral of x^2ln(x)+e^{2x}+1 is 1/3 x^3ln(x)-(x^3)/9+1/2 e^{2x}+x+C