What is the integral from 0 to 1 of x7^x ?

The integral from 0 to 1 of x7^x is (7ln(7)-6)/(ln^2(7))

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integral from 0 to 1 of x7^x

\int_{0}^{1} x 7^{x} d x

\frac{7 ln ( 7 ) - 6}{ln ^{2} ( 7 )}

Decimal Notation

2.01273 \dots

Solution steps

\int_{0}^{1} x \cdot 7^{x} d x

Apply Integration By Parts

: [\frac{7 ^{x} x}{ln ( 7 )} - \int \frac{7 ^{x}}{ln ( 7 )} d x]_{0}^{1}

= [\frac{7 ^{x} x}{ln ( 7 )} - \int \frac{7 ^{x}}{ln ( 7 )} d x]_{0}^{1}

\int \frac{7 ^{x}}{ln ( 7 )} d x = \frac{7 ^{x}}{ln ^{2} ( 7 )}

= [\frac{7 ^{x} x}{ln ( 7 )} - \frac{7 ^{x}}{ln ^{2} ( 7 )}]_{0}^{1}

Compute the boundaries:

\frac{7 ln ( 7 ) - 6}{ln ^{2} ( 7 )}

= \frac{7 ln ( 7 ) - 6}{ln ^{2} ( 7 )}

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What is the integral from 0 to 1 of x7^x ?
The integral from 0 to 1 of x7^x is (7ln(7)-6)/(ln^2(7))