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

= \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

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

x = (y - 1)^{2}, x = 13 - y^{2}

\int \frac{2 x}{e ^{2 x}} d x

x \to 2 + lim (g (x))

\frac{d}{d x} (arctan (8^{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