What is the integral of x^{x'} ?

The integral of x^{x'} is x^{x'+1}-(x^2)/2+C

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integral of x^{x'}

\int x^{x^{'}} d x

x^{x' + 1} - \frac{x ^{2}}{2} + C

Solution steps

\int x^{x^{'}} d x

Apply Integration By Parts

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

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

= x^{x^{^{'}} + 1} - \frac{x ^{2}}{2}

Add a constant to the solution

= x^{x^{^{'}} + 1} - \frac{x ^{2}}{2} + C