What is the integral from 0 to 1 of (3^{3x}-1)/(3^x) ?

The integral from 0 to 1 of (3^{3x}-1)/(3^x) is (10)/(3ln(3))

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integral from 0 to 1 of (3^{3x}-1)/(3^x)

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

\frac{10}{3 ln ( 3 )}

Decimal Notation

3.03413 \dots

Solution steps

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

Expand

\frac{3 ^{3 x} - 1}{3 ^{x}} : 9^{x} - \frac{1}{3 ^{x}}

= \int_{0}^{1} 9^{x} - \frac{1}{3 ^{x}} 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_{0}^{1} 9^{x} d x - \int_{0}^{1} \frac{1}{3 ^{x}} d x

\int_{0}^{1} 9^{x} d x = \frac{4}{ln ( 3 )}

\int_{0}^{1} \frac{1}{3 ^{x}} d x = \frac{2}{3 ln ( 3 )}

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

Simplify

\frac{4}{ln ( 3 )} - \frac{2}{3 ln ( 3 )} : \frac{10}{3 ln ( 3 )}

= \frac{10}{3 ln ( 3 )}

y = (2 x + 1) + \frac{2}{( 2 x + 1 ) ^{3}}

\int_{- 1}^{1} x (1 - x)^{2} d x

x \to 0 + lim (\frac{1}{sin ^{2} ( x )})

\frac{d}{d x} (\frac{24 x ( x ^{2} + 1 )}{( 1 - x ^{2} ) ^{4}})

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

What is the integral from 0 to 1 of (3^{3x}-1)/(3^x) ?
The integral from 0 to 1 of (3^{3x}-1)/(3^x) is (10)/(3ln(3))