What is the integral of ln(3x+5) ?

The integral of ln(3x+5) is xln(3x+5)+5/3 ln(3x+5)-x-5/3+C

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integral of ln(3x+5)

\int ln (3 x + 5) d x

x ln (3 x + 5) + \frac{5}{3} ln (3 x + 5) - x - \frac{5}{3} + C

Solution steps

\int ln (3 x + 5) 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 + 5

= \frac{1}{3} ((3 x + 5) ln (3 x + 5) - (3 x + 5))

Simplify

\frac{1}{3} ((3 x + 5) ln (3 x + 5) - (3 x + 5)) : x ln (3 x + 5) + \frac{5}{3} ln (3 x + 5) - x - \frac{5}{3}

= x ln (3 x + 5) + \frac{5}{3} ln (3 x + 5) - x - \frac{5}{3}

Add a constant to the solution

= x ln (3 x + 5) + \frac{5}{3} ln (3 x + 5) - x - \frac{5}{3} + C

f (x) = 2 x^{2} - 5 x + 3

\frac{d}{d x} ((a x^{2} + b x + c) e^{\frac{- x}{2}})

\frac{1}{3 s ^{2}} - \frac{5}{4 s ^{3}}

\frac{d}{d x} (arcsin (2 x) + arccos (2 x))

y^{^{''}} - y^{^{'}} - 6 y = 2 e^{2 t} - t^{2} - 2 t + 3

What is the integral of ln(3x+5) ?
The integral of ln(3x+5) is xln(3x+5)+5/3 ln(3x+5)-x-5/3+C