What is the integral of ln(4x+1) ?

The integral of ln(4x+1) is xln(4x+1)+1/4 ln(4x+1)-x-1/4+C

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integral of ln(4x+1)

\int ln (4 x + 1) d x

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

Solution steps

\int ln (4 x + 1) d x

Apply u-substitution

: \int ln (u) \frac{1}{4} d u

= \int ln (u) \frac{1}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int ln (u) d u

Apply Integration By Parts

: u ln (u) - \int 1 d u

= \frac{1}{4} (u ln (u) - \int 1 d u)

\int 1 d u = u

= \frac{1}{4} (u ln (u) - u)

Substitute back

u = 4 x + 1

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

Simplify

\frac{1}{4} ((4 x + 1) ln (4 x + 1) - (4 x + 1)) : x ln (4 x + 1) + \frac{1}{4} ln (4 x + 1) - x - \frac{1}{4}

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

Add a constant to the solution

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

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What is the integral of ln(4x+1) ?
The integral of ln(4x+1) is xln(4x+1)+1/4 ln(4x+1)-x-1/4+C