What is the integral of (3yln(y^2-5))/(y^2-5) ?

The integral of (3yln(y^2-5))/(y^2-5) is 3/4 ln^2(y^2-5)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (3yln(y^2-5))/(y^2-5)

\int \frac{3 y ln ( y ^{2} - 5 )}{y ^{2} - 5} d y

\frac{3}{4} ln^{2} (y^{2} - 5) + C

Solution steps

\int \frac{3 y ln ( y ^{2} - 5 )}{y ^{2} - 5} d y

Take the constant out:

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

= 3 \cdot \int \frac{y ln ( y ^{2} - 5 )}{y ^{2} - 5} d y

Apply u-substitution

: \int \frac{ln ( u - 5 )}{2 ( u - 5 )} d u

= 3 \cdot \int \frac{ln ( u - 5 )}{2 ( u - 5 )} 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 \frac{ln ( u - 5 )}{u - 5} d u

Apply u-substitution

: \int \frac{ln ( v )}{v} d v

= 3 \cdot \frac{1}{2} \cdot \int \frac{ln ( v )}{v} d v

Apply u-substitution

: \int w d w

= 3 \cdot \frac{1}{2} \cdot \int w d w

Apply the Power Rule

: \frac{w ^{2}}{2}

= 3 \cdot \frac{1}{2} \cdot \frac{w ^{2}}{2}

Substitute back

= 3 \cdot \frac{1}{2} \cdot \frac{ln ^{2} ( y ^{2} - 5 )}{2}

Simplify

3 \cdot \frac{1}{2} \cdot \frac{ln ^{2} ( y ^{2} - 5 )}{2} : \frac{3}{4} ln^{2} (y^{2} - 5)

= \frac{3}{4} ln^{2} (y^{2} - 5)

Add a constant to the solution

= \frac{3}{4} ln^{2} (y^{2} - 5) + C

\int (\frac{1}{2} sec^{2} (x) + x) d x

\int \frac{1}{2000} e^{- \frac{x}{2000}} d x

\int - 10 - 5 x^{2} d x

\int tan^{4} (2 x) sec^{4} (2 x) d x

What is the integral of (3yln(y^2-5))/(y^2-5) ?
The integral of (3yln(y^2-5))/(y^2-5) is 3/4 ln^2(y^2-5)+C