What is the integral of (x^2+6)/(xsqrt(x^2-9)) ?

The integral of (x^2+6)/(xsqrt(x^2-9)) is sqrt(x^2-9)+2arctan(1/3 sqrt(x^2-9))+C

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integral of (x^2+6)/(xsqrt(x^2-9))

\int \frac{x ^{2} + 6}{x x ^{2} - 9} d x

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

Solution steps

\int \frac{x ^{2} + 6}{x x ^{2} - 9} d x

Apply u-substitution

: \int \frac{u + 6}{2 u u - 9} d u

= \int \frac{u + 6}{2 u u - 9} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{u + 6}{u u - 9} d u

Expand

\frac{u + 6}{u u - 9} : \frac{1}{u - 9} + \frac{6}{u u - 9}

= \frac{1}{2} \cdot \int \frac{1}{u - 9} + \frac{6}{u u - 9} d u

Apply the Sum Rule:

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

= \frac{1}{2} (\int \frac{1}{u - 9} d u + \int \frac{6}{u u - 9} d u)

\int \frac{1}{u - 9} d u = 2 u - 9

\int \frac{6}{u u - 9} d u = 4 arctan (\frac{1}{3} u - 9)

= \frac{1}{2} (2 u - 9 + 4 arctan (\frac{1}{3} u - 9))

Substitute back

u = x^{2}

= \frac{1}{2} (2 x^{2} - 9 + 4 arctan (\frac{1}{3} x^{2} - 9))

Simplify

\frac{1}{2} (2 x^{2} - 9 + 4 arctan (\frac{1}{3} x^{2} - 9)) : x^{2} - 9 + 2 arctan (\frac{1}{3} x^{2} - 9)

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{e ^{x} \cdot cos ( x )}{e ^{x} - 1})

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

x \to + (- 8) lim (2)

x^{2} + x - 2, 0

\int \frac{x ^{3}}{2 x ^{4} + 1} d x

What is the integral of (x^2+6)/(xsqrt(x^2-9)) ?
The integral of (x^2+6)/(xsqrt(x^2-9)) is sqrt(x^2-9)+2arctan(1/3 sqrt(x^2-9))+C