What is the integral of ln(2x+7) ?

The integral of ln(2x+7) is xln(2x+7)+7/2 ln(2x+7)-x-7/2+C

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integral of ln(2x+7)

\int ln (2 x + 7) d x

x ln (2 x + 7) + \frac{7}{2} ln (2 x + 7) - x - \frac{7}{2} + C

Solution steps

\int ln (2 x + 7) d x

Apply u-substitution

: \int ln (u) \frac{1}{2} d u

= \int ln (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int ln (u) d u

Apply Integration By Parts

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

= \frac{1}{2} (u ln (u) - \int 1 d u)

\int 1 d u = u

= \frac{1}{2} (u ln (u) - u)

Substitute back

u = 2 x + 7

= \frac{1}{2} ((2 x + 7) ln (2 x + 7) - (2 x + 7))

Simplify

\frac{1}{2} ((2 x + 7) ln (2 x + 7) - (2 x + 7)) : x ln (2 x + 7) + \frac{7}{2} ln (2 x + 7) - x - \frac{7}{2}

= x ln (2 x + 7) + \frac{7}{2} ln (2 x + 7) - x - \frac{7}{2}

Add a constant to the solution

= x ln (2 x + 7) + \frac{7}{2} ln (2 x + 7) - x - \frac{7}{2} + C

\frac{d}{d x} ((sin (x))^{t a n (x)})

\int_{0}^{e} 2 π y (1 - ln (y)) d y

\frac{8}{x + 11}, at x = 5

n = 0 \sum \infty (\frac{3}{7})^{\frac{n}{2}}

\int_{- 1}^{1} \frac{1}{2} x^{3} - 2 x + 2 d x

What is the integral of ln(2x+7) ?
The integral of ln(2x+7) is xln(2x+7)+7/2 ln(2x+7)-x-7/2+C