What is the integral of s/(s^2+9) ?

The integral of s/(s^2+9) is 1/2 ln|s^2+9|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of s/(s^2+9)

integral

\frac{s}{s ^{2} + 9}

\frac{1}{2} ln s^{2} + 9 + C

Solution steps

\int \frac{s}{s ^{2} + 9} d s

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = s^{2} + 9

= \frac{1}{2} ln s^{2} + 9

Add a constant to the solution

= \frac{1}{2} ln s^{2} + 9 + C

\frac{1}{[ ( 1 - x ^{2} ) ( 1 + x ) ]}

\frac{1}{cos ( 2 z ) - 1}

\frac{v + 1}{v ^{2} + 1}

7 x^{6}

(x^{2} + a^{2})^{\frac{1}{2}}

What is the integral of s/(s^2+9) ?
The integral of s/(s^2+9) is 1/2 ln|s^2+9|+C