What is the integral from 1 to e of ((2+ln(x))^4)/x ?

The integral from 1 to e of ((2+ln(x))^4)/x is 211/5

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integral from 1 to e of ((2+ln(x))^4)/x

\int_{1}^{e} \frac{( 2 + ln ( x ) ) ^{4}}{x} d x

\frac{211}{5}

Decimal Notation

42.2

Solution steps

\int_{1}^{e} \frac{( 2 + ln ( x ) ) ^{4}}{x} d x

Expand

\frac{( 2 + ln ( x ) ) ^{4}}{x} : \frac{16}{x} + \frac{32 ln ( x )}{x} + \frac{24 ln ^{2} ( x )}{x} + \frac{8 ln ^{3} ( x )}{x} + \frac{ln ^{4} ( x )}{x}

= \int_{1}^{e} \frac{16}{x} + \frac{32 ln ( x )}{x} + \frac{24 ln ^{2} ( x )}{x} + \frac{8 ln ^{3} ( x )}{x} + \frac{ln ^{4} ( x )}{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_{1}^{e} \frac{16}{x} d x + \int_{1}^{e} \frac{32 ln ( x )}{x} d x + \int_{1}^{e} \frac{24 ln ^{2} ( x )}{x} d x + \int_{1}^{e} \frac{8 ln ^{3} ( x )}{x} d x + \int_{1}^{e} \frac{ln ^{4} ( x )}{x} d x

\int_{1}^{e} \frac{16}{x} d x = 16

\int_{1}^{e} \frac{32 ln ( x )}{x} d x = 16

\int_{1}^{e} \frac{24 ln ^{2} ( x )}{x} d x = 8

\int_{1}^{e} \frac{8 ln ^{3} ( x )}{x} d x = 2

\int_{1}^{e} \frac{ln ^{4} ( x )}{x} d x = \frac{1}{5}

= 16 + 16 + 8 + 2 + \frac{1}{5}

Simplify

16 + 16 + 8 + 2 + \frac{1}{5} : \frac{211}{5}

= \frac{211}{5}

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What is the integral from 1 to e of ((2+ln(x))^4)/x ?
The integral from 1 to e of ((2+ln(x))^4)/x is 211/5