What is the integral of ln(4x+9) ?

The integral of ln(4x+9) is xln(4x+9)+9/4 ln(4x+9)-x-9/4+C

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integral of ln(4x+9)

\int ln (4 x + 9) d x

x ln (4 x + 9) + \frac{9}{4} ln (4 x + 9) - x - \frac{9}{4} + C

Solution steps

\int ln (4 x + 9) d x

Apply u-substitution

: \int ln (u) \frac{1}{4} d u

= \int ln (u) \frac{1}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int ln (u) d u

Apply Integration By Parts

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

= \frac{1}{4} (u ln (u) - \int 1 d u)

\int 1 d u = u

= \frac{1}{4} (u ln (u) - u)

Substitute back

u = 4 x + 9

= \frac{1}{4} ((4 x + 9) ln (4 x + 9) - (4 x + 9))

Simplify

\frac{1}{4} ((4 x + 9) ln (4 x + 9) - (4 x + 9)) : x ln (4 x + 9) + \frac{9}{4} ln (4 x + 9) - x - \frac{9}{4}

= x ln (4 x + 9) + \frac{9}{4} ln (4 x + 9) - x - \frac{9}{4}

Add a constant to the solution

= x ln (4 x + 9) + \frac{9}{4} ln (4 x + 9) - x - \frac{9}{4} + C

\frac{\partial}{\partial x} (\frac{1}{( x ^{2} + y ^{2} + z ^{2} )})

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

\frac{d y}{d t} = \frac{3}{t} y + t^{5}

x \to 8 lim (- 6 x)

\frac{\partial}{\partial x} (- \frac{x}{2 x ^{2} + y ^{2}})

What is the integral of ln(4x+9) ?
The integral of ln(4x+9) is xln(4x+9)+9/4 ln(4x+9)-x-9/4+C