What is the integral from 1 to 5 of (ln(y))/y ?

The integral from 1 to 5 of (ln(y))/y is (ln^2(5))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 5 of (ln(y))/y

\int_{1}^{5} \frac{ln ( y )}{y} d y

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

Decimal Notation

1.29514 \dots

Solution steps

\int_{1}^{5} \frac{ln ( y )}{y} d y

Apply u-substitution

: \int_{0}^{l n (5)} u d u

= \int_{0}^{l n (5)} u d u

Apply the Power Rule

: [\frac{u ^{2}}{2}]_{0}^{l n (5)}

= [\frac{u ^{2}}{2}]_{0}^{l n (5)}

Compute the boundaries:

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

= \frac{ln ^{2} ( 5 )}{2}

\int_{0}^{π} 1 - sin (x) d x

\int_{- 1}^{3} (x^{2} - 1)^{2} d x

\int_{0}^{3} (3 - 2 x + 3 x^{2}) d x

\int_{- 2}^{1} ∣ x - 1 ∣ d x

\int_{0}^{1} 9 cos (x^{2}) d x

What is the integral from 1 to 5 of (ln(y))/y ?
The integral from 1 to 5 of (ln(y))/y is (ln^2(5))/2