What is the integral of ln(8x+1) ?

The integral of ln(8x+1) is xln(8x+1)+1/8 ln(8x+1)-x-1/8+C

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integral of ln(8x+1)

\int ln (8 x + 1) d x

x ln (8 x + 1) + \frac{1}{8} ln (8 x + 1) - x - \frac{1}{8} + C

Solution steps

\int ln (8 x + 1) d x

Apply u-substitution

= \int ln (u) \frac{1}{8} d u

Take the constant out:

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

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

Apply Integration By Parts

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

\int 1 d u = u

= \frac{1}{8} (u ln (u) - u)

Substitute back

u = 8 x + 1

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

Simplify

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

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

Add a constant to the solution

= x ln (8 x + 1) + \frac{1}{8} ln (8 x + 1) - x - \frac{1}{8} + C

What is the integral of ln(8x+1) ?
The integral of ln(8x+1) is xln(8x+1)+1/8 ln(8x+1)-x-1/8+C