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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to e of (3ln(x^2))/x

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

3

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{1}^{e^{2}} \frac{ln ( u )}{2 u} d u

= 3 \cdot \int_{1}^{e^{2}} \frac{ln ( u )}{2 u} d u

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int_{1}^{e^{2}} \frac{ln ( u )}{u} d u

Apply u-substitution

: \int_{0}^{2} v d v

= 3 \cdot \frac{1}{2} \cdot \int_{0}^{2} v d v

Apply the Power Rule

: [\frac{v ^{2}}{2}]_{0}^{2}

= 3 \cdot \frac{1}{2} [\frac{v ^{2}}{2}]_{0}^{2}

Simplify

3 \cdot \frac{1}{2} [\frac{v ^{2}}{2}]_{0}^{2} : \frac{3}{2} [\frac{v ^{2}}{2}]_{0}^{2}

= \frac{3}{2} [\frac{v ^{2}}{2}]_{0}^{2}

Compute the boundaries:

2

= \frac{3}{2} \cdot 2

Simplify

= 3

\int_{0}^{0.4} 1 + x^{2} d x

\int_{- 2 π}^{4 π} cos^{2} (x) d x

\int_{1}^{2} 2 sin (x) + cos (x) d x

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

\int_{0}^{4} (x - 1) d x

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