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

The integral of x(ln(x)+e^x) is 1/2 x^2ln(x)-x^2+(x^2)/4+(x^2)/2+e^xx-e^x+C

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integral of x(ln(x)+e^x)

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

\frac{1}{2} x^{2} ln (x) - x^{2} + \frac{x ^{2}}{4} + \frac{x ^{2}}{2} + e^{x} x - e^{x} + C

Solution steps

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

Apply Integration By Parts

: x (x ln (x) - x + e^{x}) - \int x ln (x) - x + e^{x} d x

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

\int x ln (x) - x + e^{x} d x = \frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4} - \frac{x ^{2}}{2} + e^{x}

= x (x ln (x) - x + e^{x}) - (\frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4} - \frac{x ^{2}}{2} + e^{x})

Simplify

x (x ln (x) - x + e^{x}) - (\frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4} - \frac{x ^{2}}{2} + e^{x}) : \frac{1}{2} x^{2} ln (x) - x^{2} + \frac{x ^{2}}{4} + \frac{x ^{2}}{2} + e^{x} x - e^{x}

= \frac{1}{2} x^{2} ln (x) - x^{2} + \frac{x ^{2}}{4} + \frac{x ^{2}}{2} + e^{x} x - e^{x}

Add a constant to the solution

= \frac{1}{2} x^{2} ln (x) - x^{2} + \frac{x ^{2}}{4} + \frac{x ^{2}}{2} + e^{x} x - e^{x} + C

f (x) = 1 - x^{2}, at x = 2

\int \frac{21}{25 + 4 r ^{2}} d r

\frac{1}{x} - 1

\frac{\partial}{\partial x} (2 y - 3 x + 5)

\int (e^{6 x} + e^{- 9 x}) d x

What is the integral of x(ln(x)+e^x) ?
The integral of x(ln(x)+e^x) is 1/2 x^2ln(x)-x^2+(x^2)/4+(x^2)/2+e^xx-e^x+C