What is the integral of (x+1)/(sqrt(x^2-9)) ?

The integral of (x+1)/(sqrt(x^2-9)) is sqrt(x^2-9)+ln(1/3 |sqrt(x^2-9)+x|)+C

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integral of (x+1)/(sqrt(x^2-9))

\int \frac{x + 1}{x ^{2} - 9} d x

x^{2} - 9 + ln (\frac{1}{3} x^{2} - 9 + x) + C

Solution steps

\int \frac{x + 1}{x ^{2} - 9} d x

Expand

\frac{x + 1}{x ^{2} - 9} : \frac{x}{x ^{2} - 9} + \frac{1}{x ^{2} - 9}

= \int \frac{x}{x ^{2} - 9} + \frac{1}{x ^{2} - 9} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x}{x ^{2} - 9} d x + \int \frac{1}{x ^{2} - 9} d x

\int \frac{x}{x ^{2} - 9} d x = x^{2} - 9

\int \frac{1}{x ^{2} - 9} d x = ln (\frac{1}{3} x^{2} - 9 + x)

= x^{2} - 9 + ln (\frac{1}{3} x^{2} - 9 + x)

Add a constant to the solution

= x^{2} - 9 + ln (\frac{1}{3} x^{2} - 9 + x) + C

\frac{\partial}{\partial x} (- 3 x^{2} y^{3} sin (x^{3} y^{3}))

f (0) = \frac{x}{x ^{2} + 1}

\int e^{- 0.01 t} d t

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

\frac{\partial}{\partial y} (2 x - 3 yz)

What is the integral of (x+1)/(sqrt(x^2-9)) ?
The integral of (x+1)/(sqrt(x^2-9)) is sqrt(x^2-9)+ln(1/3 |sqrt(x^2-9)+x|)+C