What is the integral from 0 to 1 of xpi^x ?

The integral from 0 to 1 of xpi^x is pi/(ln(pi))+(-pi+1)/(ln^2(pi))

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integral from 0 to 1 of xpi^x

\int_{0}^{1} x π^{x} d x

\frac{π}{ln ( π )} + \frac{- π + 1}{ln ^{2} ( π )}

Decimal Notation

1.11010 \dots

Solution steps

\int_{0}^{1} x π^{x} d x

Apply Integration By Parts

: [\frac{π ^{x} x}{ln ( π )} - \int \frac{π ^{x}}{ln ( π )} d x]_{0}^{1}

= [\frac{π ^{x} x}{ln ( π )} - \int \frac{π ^{x}}{ln ( π )} d x]_{0}^{1}

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

= [\frac{π ^{x} x}{ln ( π )} - \frac{1}{ln ^{2} ( π )} π^{x}]_{0}^{1}

Compute the boundaries:

\frac{π}{ln ( π )} - \frac{π}{ln ^{2} ( π )} + \frac{1}{ln ^{2} ( π )}

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

Simplify

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

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What is the integral from 0 to 1 of xpi^x ?
The integral from 0 to 1 of xpi^x is pi/(ln(pi))+(-pi+1)/(ln^2(pi))