What is the integral of (7ln(x+9))/(x^2) ?

The integral of (7ln(x+9))/(x^2) is 7(-(ln(x+9))/x+1/9 ln|x|-1/9 ln|x+9|)+C

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integral of (7ln(x+9))/(x^2)

\int \frac{7 ln ( x + 9 )}{x ^{2}} d x

7 (- \frac{ln ( x + 9 )}{x} + \frac{1}{9} ln ∣ x ∣ - \frac{1}{9} ln ∣ x + 9 ∣) + C

Solution steps

\int \frac{7 ln ( x + 9 )}{x ^{2}} d x

Take the constant out:

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

= 7 \cdot \int \frac{ln ( x + 9 )}{x ^{2}} d x

Apply Integration By Parts

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

= 7 (- \frac{ln ( x + 9 )}{x} - \int - \frac{1}{x ( x + 9 )} d x)

\int - \frac{1}{x ( x + 9 )} d x = - \frac{1}{9} ln ∣ x ∣ + \frac{1}{9} ln ∣ x + 9 ∣

= 7 (- \frac{ln ( x + 9 )}{x} - (- \frac{1}{9} ln ∣ x ∣ + \frac{1}{9} ln ∣ x + 9 ∣))

Simplify

= 7 (- \frac{ln ( x + 9 )}{x} + \frac{1}{9} ln ∣ x ∣ - \frac{1}{9} ln ∣ x + 9 ∣)

Add a constant to the solution

= 7 (- \frac{ln ( x + 9 )}{x} + \frac{1}{9} ln ∣ x ∣ - \frac{1}{9} ln ∣ x + 9 ∣) + C

x \to \infty lim (1 + e^{x})

x \to 2 lim (\frac{( x - 2 )}{( x - 1 - 3 - x )})

f (x) = x - x^{3}, at x = 1

5 x y

f (x) = (5 x + 2)^{\frac{1}{3}}, at x = 2

What is the integral of (7ln(x+9))/(x^2) ?
The integral of (7ln(x+9))/(x^2) is 7(-(ln(x+9))/x+1/9 ln|x|-1/9 ln|x+9|)+C