What is the integral of 9/(x^{-1)+3} ?

The integral of 9/(x^{-1)+3} is 1+3x-ln|1+3x|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 9/(x^{-1)+3}

\int \frac{9}{x ^{- 1} + 3} d x

1 + 3 x - ln ∣ 1 + 3 x ∣ + C

Solution steps

\int \frac{9}{x ^{- 1} + 3} d x

Take the constant out:

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

= 9 \cdot \int \frac{1}{x ^{- 1} + 3} d x

Simplify

\frac{1}{x ^{- 1} + 3} : \frac{x}{1 + 3 x}

= 9 \cdot \int \frac{x}{1 + 3 x} d x

Apply u-substitution

: \int \frac{u - 1}{9 u} d u

= 9 \cdot \int \frac{u - 1}{9 u} d u

Take the constant out:

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

= 9 \cdot \frac{1}{9} \cdot \int \frac{u - 1}{u} d u

Expand

\frac{u - 1}{u} : 1 - \frac{1}{u}

= 9 \cdot \frac{1}{9} \cdot \int 1 - \frac{1}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 9 \cdot \frac{1}{9} (\int 1 d u - \int \frac{1}{u} d u)

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= 9 \cdot \frac{1}{9} (u - ln ∣ u ∣)

Substitute back

u = 1 + 3 x

= 9 \cdot \frac{1}{9} (1 + 3 x - ln ∣ 1 + 3 x ∣)

Simplify

9 \cdot \frac{1}{9} (1 + 3 x - ln ∣ 1 + 3 x ∣) : 1 + 3 x - ln ∣ 1 + 3 x ∣

= 1 + 3 x - ln ∣ 1 + 3 x ∣

Add a constant to the solution

= 1 + 3 x - ln ∣ 1 + 3 x ∣ + C

\frac{\partial}{\partial x} (y cos (x^{2} y))

(8 \frac{1}{7}, 9), (4, 3 \frac{1}{5})

y = ln (x^{2} - 5 x + 1), (5, 0)

\int_{0}^{\frac{π}{3}} x sin (6 x) d x

\int (1 + x)^{0} d x

What is the integral of 9/(x^{-1)+3} ?
The integral of 9/(x^{-1)+3} is 1+3x-ln|1+3x|+C