What is the integral of (x^3)/(sqrt(x^2-9)) ?

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

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integral of (x^3)/(sqrt(x^2-9))

\int \frac{x ^{3}}{x ^{2} - 9} d x

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

Solution steps

\int \frac{x ^{3}}{x ^{2} - 9} d x

Apply u-substitution

= \int u^{2} + 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

= \int u^{2} d u + \int 9 d u

\int u^{2} d u = \frac{u ^{3}}{3}

\int 9 d u = 9 u

= \frac{u ^{3}}{3} + 9 u

Substitute back

u = x^{2} - 9

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

Simplify

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

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

Add a constant to the solution

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

t a y l or sin (2 x), 0

\frac{d}{d t} (\frac{e ^{- t}}{1 + t ^{2}})

d er i v a t i v e arcsec (x) \cdot arccos (x)

\int t e^{t + 2} d t

\int \frac{d m}{m ^{3} + m ^{2}} d x

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