What is the integral of ln(7x+1) ?

The integral of ln(7x+1) is xln(7x+1)+1/7 ln(7x+1)-x-1/7+C

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integral of ln(7x+1)

\int ln (7 x + 1) d x

x ln (7 x + 1) + \frac{1}{7} ln (7 x + 1) - x - \frac{1}{7} + C

Solution steps

\int ln (7 x + 1) d x

Apply u-substitution

: \int ln (u) \frac{1}{7} d u

= \int ln (u) \frac{1}{7} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int ln (u) d u

Apply Integration By Parts

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

= \frac{1}{7} (u ln (u) - \int 1 d u)

\int 1 d u = u

= \frac{1}{7} (u ln (u) - u)

Substitute back

u = 7 x + 1

= \frac{1}{7} ((7 x + 1) ln (7 x + 1) - (7 x + 1))

Simplify

\frac{1}{7} ((7 x + 1) ln (7 x + 1) - (7 x + 1)) : x ln (7 x + 1) + \frac{1}{7} ln (7 x + 1) - x - \frac{1}{7}

= x ln (7 x + 1) + \frac{1}{7} ln (7 x + 1) - x - \frac{1}{7}

Add a constant to the solution

= x ln (7 x + 1) + \frac{1}{7} ln (7 x + 1) - x - \frac{1}{7} + C

\frac{d}{d x} ((3 x^{5} - 1) (2 - x^{4}))

y = x x + 1

\int 2 x^{3} x^{2} - 1 d x

\int 7 e^{- 0.2 x} d x

\frac{d}{d x} (\frac{6 x}{9 - tan ( x )})

What is the integral of ln(7x+1) ?
The integral of ln(7x+1) is xln(7x+1)+1/7 ln(7x+1)-x-1/7+C