What is the integral of (4x^3)/(sqrt(x^2+9)) ?

The integral of (4x^3)/(sqrt(x^2+9)) is 4(1/3 (x^2+9)^{3/2}-9sqrt(x^2+9))+C

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integral of (4x^3)/(sqrt(x^2+9))

\int \frac{4 x ^{3}}{x ^{2} + 9} d x

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

Solution steps

\int \frac{4 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

= 4 \cdot \int \frac{x ^{3}}{x ^{2} + 9} d x

Apply u-substitution

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

= 4 \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

= 4 (\int u^{2} d u - \int 9 d u)

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

\int 9 d u = 9 u

= 4 (\frac{u ^{3}}{3} - 9 u)

Substitute back

u = x^{2} + 9

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

Simplify

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

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

Add a constant to the solution

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

f (x) = 2 x^{2} - 2 x - 2, at x = 1

f^{^{'}} (x) = cos (cos (x))

n = 0 \sum \infty n + 1 (\frac{1}{9})^{n}

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

\int x (2 x^{2} - 4)^{3} d x

What is the integral of (4x^3)/(sqrt(x^2+9)) ?
The integral of (4x^3)/(sqrt(x^2+9)) is 4(1/3 (x^2+9)^{3/2}-9sqrt(x^2+9))+C