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

The integral from 1 to e of ((1+ln(x))^2)/x is 7/3

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

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

\frac{7}{3}

Decimal Notation

2.33333 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{2} u^{2} d u

= \int_{1}^{2} u^{2} d u

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{1}^{2}

= [\frac{u ^{3}}{3}]_{1}^{2}

Compute the boundaries:

\frac{7}{3}

= \frac{7}{3}

\int \frac{tan ( ln ( x ) )}{x} d x

(3 x^{3} + x^{- 3}) (x + 3) (x^{2} - 5)

f^{^{'}} (x) = 2 x^{3} - x

y = c e^{\frac{- x ^{3}}{3}} + 1

\int_{1}^{5} 6250 π x d x

What is the integral from 1 to e of ((1+ln(x))^2)/x ?
The integral from 1 to e of ((1+ln(x))^2)/x is 7/3