What is the integral from 0 to 1 of-3x5^x ?

The integral from 0 to 1 of-3x5^x is -(3(5ln(5)-4))/(ln^2(5))

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integral from 0 to 1 of-3x5^x

\int_{0}^{1} - 3 x 5^{x} d x

- \frac{3 ( 5 ln ( 5 ) - 4 )}{ln ^{2} ( 5 )}

Decimal Notation

- 4.68733 \dots

Solution steps

\int_{0}^{1} - 3 x \cdot 5^{x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 3 \cdot \int_{0}^{1} x \cdot 5^{x} d x

Apply Integration By Parts

: [\frac{5 ^{x} x}{ln ( 5 )} - \int \frac{5 ^{x}}{ln ( 5 )} d x]_{0}^{1}

= - 3 [\frac{5 ^{x} x}{ln ( 5 )} - \int \frac{5 ^{x}}{ln ( 5 )} d x]_{0}^{1}

\int \frac{5 ^{x}}{ln ( 5 )} d x = \frac{5 ^{x}}{ln ^{2} ( 5 )}

= - 3 [\frac{5 ^{x} x}{ln ( 5 )} - \frac{5 ^{x}}{ln ^{2} ( 5 )}]_{0}^{1}

Compute the boundaries:

\frac{5 ln ( 5 ) - 4}{ln ^{2} ( 5 )}

= - 3 \cdot \frac{5 ln ( 5 ) - 4}{ln ^{2} ( 5 )}

Simplify

= - \frac{3 ( 5 ln ( 5 ) - 4 )}{ln ^{2} ( 5 )}

\frac{d}{d x} (ln (7 x^{5}))

\frac{d}{d x} (\frac{x e ^{x} - 9 x}{e ^{x}})

9 x^{- 4} - 9 x^{- 1}

\int_{- \infty}^{1} 5^{x} d x

x^{^{''}} + 9 x - 27 = 0

What is the integral from 0 to 1 of-3x5^x ?
The integral from 0 to 1 of-3x5^x is -(3(5ln(5)-4))/(ln^2(5))