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

The integral of (ln(x+9))/(x+9) is (ln^2(x+9))/2+C

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integral of (ln(x+9))/(x+9)

\int \frac{ln ( x + 9 )}{x + 9} d x

\frac{ln ^{2} ( x + 9 )}{2} + C

Solution steps

\int \frac{ln ( x + 9 )}{x + 9} d x

Apply u-substitution

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

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

Apply u-substitution

: \int v d v

= \int v d v

Apply the Power Rule

: \frac{v ^{2}}{2}

= \frac{v ^{2}}{2}

Substitute back

= \frac{ln ^{2} ( x + 9 )}{2}

Add a constant to the solution

= \frac{ln ^{2} ( x + 9 )}{2} + C

\int_{- 1}^{9} (9 - x^{2}) - (- 8 x) d x

y^{^{'}} = - 3 y^{2}

\frac{d}{d x} (\frac{1}{12} (x + 3)^{3} + (x + 3)^{- 1})

\int_{1}^{4} t ln (\frac{t}{2}) d t

\int \frac{4 x}{1 - 16 x ^{4}} d x

What is the integral of (ln(x+9))/(x+9) ?
The integral of (ln(x+9))/(x+9) is (ln^2(x+9))/2+C