What is the integral of ((t+9))/(7t(t+9)-6(t+9)) ?

The integral of ((t+9))/(7t(t+9)-6(t+9)) is 1/7 ln|7t-6|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((t+9))/(7t(t+9)-6(t+9))

\int \frac{( t + 9 )}{7 t ( t + 9 ) - 6 ( t + 9 )} d t

\frac{1}{7} ln ∣ 7 t - 6 ∣ + C

Solution steps

\int \frac{t + 9}{7 t ( t + 9 ) - 6 ( t + 9 )} d t

Simplify

\frac{t + 9}{7 t ( t + 9 ) - 6 ( t + 9 )} : \frac{1}{7 t - 6}

= \int \frac{1}{7 t - 6} d t

Apply u-substitution

: \int \frac{1}{7 u} d u

= \int \frac{1}{7 u} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{7} ln ∣ u ∣

Substitute back

u = 7 t - 6

= \frac{1}{7} ln ∣ 7 t - 6 ∣

Add a constant to the solution

= \frac{1}{7} ln ∣ 7 t - 6 ∣ + C

\int x^{- 2} cos (\frac{π}{x}) d x

\int 2 x e^{x^{2} + 2} d x

\int cot (θ) csc^{2} (θ) d θ

\int sin (6 x - 2) d x

\int (3 x - 2)^{5} d x

What is the integral of ((t+9))/(7t(t+9)-6(t+9)) ?
The integral of ((t+9))/(7t(t+9)-6(t+9)) is 1/7 ln|7t-6|+C