What is the integral of (6x^3)/(sqrt(x^2+9)) ?

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

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integral of (6x^3)/(sqrt(x^2+9))

\int \frac{6 x ^{3}}{x ^{2} + 9} d x

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

Solution steps

\int \frac{6 x ^{3}}{x ^{2} + 9} d x

Take the constant out:

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

= 6 \cdot \int \frac{x ^{3}}{x ^{2} + 9} d x

Apply u-substitution

: \int u^{2} - 9 d u

= 6 \cdot \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

= 6 (\int u^{2} d u - \int 9 d u)

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

\int 9 d u = 9 u

= 6 (\frac{u ^{3}}{3} - 9 u)

Substitute back

u = x^{2} + 9

= 6 (\frac{( x ^{2} + 9 ) ^{3}}{3} - 9 x^{2} + 9)

Simplify

6 (\frac{( x ^{2} + 9 ) ^{3}}{3} - 9 x^{2} + 9) : 6 (\frac{1}{3} (x^{2} + 9)^{\frac{3}{2}} - 9 x^{2} + 9)

= 6 (\frac{1}{3} (x^{2} + 9)^{\frac{3}{2}} - 9 x^{2} + 9)

Add a constant to the solution

= 6 (\frac{1}{3} (x^{2} + 9)^{\frac{3}{2}} - 9 x^{2} + 9) + C

t \to - 1 lim (t^{2} - 1)

\int sin (5 x) \cdot cos (2 x) d x

f (t) = ln (t^{9} + 5)

y = 4 x - x^{2}, (1, 3)

\frac{d y}{d x} + 8 y = 10

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