What is the integral of 8/(x^{-1)+4} ?

The integral of 8/(x^{-1)+4} is 1/2 (1+4x-ln|1+4x|)+C

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integral of 8/(x^{-1)+4}

\int \frac{8}{x ^{- 1} + 4} d x

\frac{1}{2} (1 + 4 x - ln ∣ 1 + 4 x ∣) + C

Solution steps

\int \frac{8}{x ^{- 1} + 4} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{x ^{- 1} + 4} d x

Simplify

\frac{1}{x ^{- 1} + 4} : \frac{x}{1 + 4 x}

= 8 \cdot \int \frac{x}{1 + 4 x} d x

Apply u-substitution

: \int \frac{u - 1}{16 u} d u

= 8 \cdot \int \frac{u - 1}{16 u} d u

Take the constant out:

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

= 8 \cdot \frac{1}{16} \cdot \int \frac{u - 1}{u} d u

Expand

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

= 8 \cdot \frac{1}{16} \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

= 8 \cdot \frac{1}{16} (\int 1 d u - \int \frac{1}{u} d u)

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= 8 \cdot \frac{1}{16} (u - ln ∣ u ∣)

Substitute back

u = 1 + 4 x

= 8 \cdot \frac{1}{16} (1 + 4 x - ln ∣ 1 + 4 x ∣)

Simplify

8 \cdot \frac{1}{16} (1 + 4 x - ln ∣ 1 + 4 x ∣) : \frac{1}{2} (1 + 4 x - ln ∣ 1 + 4 x ∣)

= \frac{1}{2} (1 + 4 x - ln ∣ 1 + 4 x ∣)

Add a constant to the solution

= \frac{1}{2} (1 + 4 x - ln ∣ 1 + 4 x ∣) + C

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What is the integral of 8/(x^{-1)+4} ?
The integral of 8/(x^{-1)+4} is 1/2 (1+4x-ln|1+4x|)+C