What is the integral of (8x)/(8x+3) ?

The integral of (8x)/(8x+3) is x-3/8 ln|8x+3|+C

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integral of (8x)/(8x+3)

\int \frac{8 x}{8 x + 3} d x

x - \frac{3}{8} ln ∣ 8 x + 3 ∣ + C

Solution steps

\int \frac{8 x}{8 x + 3} d x

Take the constant out:

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

= 8 \cdot \int \frac{x}{8 x + 3} d x

Long division

\frac{x}{8 x + 3} : \frac{1}{8} - \frac{3}{8 ( 8 x + 3 )}

= 8 \cdot \int \frac{1}{8} - \frac{3}{8 ( 8 x + 3 )} d x

Apply the Sum Rule:

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

= 8 (\int \frac{1}{8} d x - \int \frac{3}{8 ( 8 x + 3 )} d x)

\int \frac{1}{8} d x = \frac{1}{8} x

\int \frac{3}{8 ( 8 x + 3 )} d x = \frac{3}{64} ln ∣ 8 x + 3 ∣

= 8 (\frac{1}{8} x - \frac{3}{64} ln ∣ 8 x + 3 ∣)

Simplify

8 (\frac{1}{8} x - \frac{3}{64} ln ∣ 8 x + 3 ∣) : x - \frac{3}{8} ln ∣ 8 x + 3 ∣

= x - \frac{3}{8} ln ∣ 8 x + 3 ∣

Add a constant to the solution

= x - \frac{3}{8} ln ∣ 8 x + 3 ∣ + C

\frac{d}{d x} ((7 - 3 x^{3})^{2})

\int_{0}^{2} x e^{- x} d x

x \to 0 lim ((x - 2) e^{- \frac{1}{x}})

x (\frac{d y}{d x}) = 2 x + 2 y

\int \frac{1}{1 + 29 x} d x

What is the integral of (8x)/(8x+3) ?
The integral of (8x)/(8x+3) is x-3/8 ln|8x+3|+C