What is the integral of ln(3x+4) ?

The integral of ln(3x+4) is xln(3x+4)+4/3 ln(3x+4)-x-4/3+C

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integral of ln(3x+4)

\int ln (3 x + 4) d x

x ln (3 x + 4) + \frac{4}{3} ln (3 x + 4) - x - \frac{4}{3} + C

Solution steps

\int ln (3 x + 4) d x

Apply u-substitution

: \int ln (u) \frac{1}{3} d u

= \int ln (u) \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int ln (u) d u

Apply Integration By Parts

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

= \frac{1}{3} (u ln (u) - \int 1 d u)

\int 1 d u = u

= \frac{1}{3} (u ln (u) - u)

Substitute back

u = 3 x + 4

= \frac{1}{3} ((3 x + 4) ln (3 x + 4) - (3 x + 4))

Simplify

\frac{1}{3} ((3 x + 4) ln (3 x + 4) - (3 x + 4)) : x ln (3 x + 4) + \frac{4}{3} ln (3 x + 4) - x - \frac{4}{3}

= x ln (3 x + 4) + \frac{4}{3} ln (3 x + 4) - x - \frac{4}{3}

Add a constant to the solution

= x ln (3 x + 4) + \frac{4}{3} ln (3 x + 4) - x - \frac{4}{3} + C

\frac{d y}{d x} = \frac{y ^{2}}{x - 1}

\frac{\partial}{\partial y} (x^{2} - 3 x y + y^{2})

f (x) = 2 x^{2} - x

\int \frac{1}{t ^{2} - 2 t + 26} d t

y^{^{'}} + y^{2} = 0

What is the integral of ln(3x+4) ?
The integral of ln(3x+4) is xln(3x+4)+4/3 ln(3x+4)-x-4/3+C