What is the integral of 5/(x^{-1)+5} ?

The integral of 5/(x^{-1)+5} is 1/5+x-1/5 ln|1+5x|+C

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integral of 5/(x^{-1)+5}

\int \frac{5}{x ^{- 1} + 5} d x

\frac{1}{5} + x - \frac{1}{5} ln ∣ 1 + 5 x ∣ + C

Solution steps

\int \frac{5}{x ^{- 1} + 5} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{x ^{- 1} + 5} d x

Simplify

\frac{1}{x ^{- 1} + 5} : \frac{x}{1 + 5 x}

= 5 \cdot \int \frac{x}{1 + 5 x} d x

Apply u-substitution

: \int \frac{u - 1}{25 u} d u

= 5 \cdot \int \frac{u - 1}{25 u} d u

Take the constant out:

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

= 5 \cdot \frac{1}{25} \cdot \int \frac{u - 1}{u} d u

Expand

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

= 5 \cdot \frac{1}{25} \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

= 5 \cdot \frac{1}{25} (\int 1 d u - \int \frac{1}{u} d u)

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= 5 \cdot \frac{1}{25} (u - ln ∣ u ∣)

Substitute back

u = 1 + 5 x

= 5 \cdot \frac{1}{25} (1 + 5 x - ln ∣ 1 + 5 x ∣)

Simplify

5 \cdot \frac{1}{25} (1 + 5 x - ln ∣ 1 + 5 x ∣) : \frac{1}{5} + x - \frac{1}{5} ln ∣ 1 + 5 x ∣

= \frac{1}{5} + x - \frac{1}{5} ln ∣ 1 + 5 x ∣

Add a constant to the solution

= \frac{1}{5} + x - \frac{1}{5} ln ∣ 1 + 5 x ∣ + C

x \to \infty lim (\frac{1}{x} - 3)

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What is the integral of 5/(x^{-1)+5} ?
The integral of 5/(x^{-1)+5} is 1/5+x-1/5 ln|1+5x|+C