What is the integral of xln(x)+1 ?

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

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integral of xln(x)+1

\int x ln (x) + 1 d x

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

Solution steps

\int x ln (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 ln (x) d x + \int 1 d x

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

\int 1 d x = x

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

Add a constant to the solution

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

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