What is the integral of 3ln(2x+1) ?

The integral of 3ln(2x+1) is 3xln(2x+1)+3/2 ln(2x+1)-3x-3/2+C

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integral of 3ln(2x+1)

\int 3 ln (2 x + 1) d x

3 x ln (2 x + 1) + \frac{3}{2} ln (2 x + 1) - 3 x - \frac{3}{2} + C

Solution steps

\int 3 ln (2 x + 1) d x

Take the constant out:

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

= 3 \cdot \int ln (2 x + 1) d x

Apply u-substitution

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

= 3 \cdot \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

= 3 \cdot \frac{1}{2} \cdot \int ln (u) d u

Apply Integration By Parts

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

= 3 \cdot \frac{1}{2} (u ln (u) - \int 1 d u)

\int 1 d u = u

= 3 \cdot \frac{1}{2} (u ln (u) - u)

Substitute back

u = 2 x + 1

= 3 \cdot \frac{1}{2} ((2 x + 1) ln (2 x + 1) - (2 x + 1))

Simplify

3 \cdot \frac{1}{2} ((2 x + 1) ln (2 x + 1) - (2 x + 1)) : 3 x ln (2 x + 1) + \frac{3}{2} ln (2 x + 1) - 3 x - \frac{3}{2}

= 3 x ln (2 x + 1) + \frac{3}{2} ln (2 x + 1) - 3 x - \frac{3}{2}

Add a constant to the solution

= 3 x ln (2 x + 1) + \frac{3}{2} ln (2 x + 1) - 3 x - \frac{3}{2} + C

\int \frac{sin ( \frac{7}{x} )}{x ^{2}} d x

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

\frac{d}{d x} (x e^{- x} - e^{- 3})

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

What is the integral of 3ln(2x+1) ?
The integral of 3ln(2x+1) is 3xln(2x+1)+3/2 ln(2x+1)-3x-3/2+C