What is the integral of ln(5x+1) ?

The integral of ln(5x+1) is xln(5x+1)+1/5 ln(5x+1)-x-1/5+C

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integral of ln(5x+1)

\int ln (5 x + 1) d x

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

Solution steps

\int ln (5 x + 1) d x

Apply u-substitution

: \int ln (u) \frac{1}{5} d u

= \int ln (u) \frac{1}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int ln (u) d u

Apply Integration By Parts

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

= \frac{1}{5} (u ln (u) - \int 1 d u)

\int 1 d u = u

= \frac{1}{5} (u ln (u) - u)

Substitute back

u = 5 x + 1

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{5 - 4 x}{4 - 4 x} d x

y = x^{3} e^{- 8 x}

\frac{\partial}{\partial y} (\frac{9 y}{x ^{3}})

\int 0.5 e^{- x} d x

\int \frac{4}{( x ^{2} + 1 ) ^{\frac{3}{2}}} d x

What is the integral of ln(5x+1) ?
The integral of ln(5x+1) is xln(5x+1)+1/5 ln(5x+1)-x-1/5+C