What is the integral of (3x^3)/(sqrt(x^2+16)) ?

The integral of (3x^3)/(sqrt(x^2+16)) is (x^2+16)^{3/2}-48sqrt(x^2+16)+C

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integral of (3x^3)/(sqrt(x^2+16))

\int \frac{3 x ^{3}}{x ^{2} + 16} d x

(x^{2} + 16)^{\frac{3}{2}} - 48 x^{2} + 16 + C

Solution steps

\int \frac{3 x ^{3}}{x ^{2} + 16} d x

Take the constant out:

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

= 3 \cdot \int \frac{x ^{3}}{x ^{2} + 16} d x

Apply u-substitution

: \int u^{2} - 16 d u

= 3 \cdot \int u^{2} - 16 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

= 3 (\int u^{2} d u - \int 16 d u)

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

\int 16 d u = 16 u

= 3 (\frac{u ^{3}}{3} - 16 u)

Substitute back

u = x^{2} + 16

= 3 (\frac{( x ^{2} + 16 ) ^{3}}{3} - 16 x^{2} + 16)

Simplify

3 (\frac{( x ^{2} + 16 ) ^{3}}{3} - 16 x^{2} + 16) : (x^{2} + 16)^{\frac{3}{2}} - 48 x^{2} + 16

= (x^{2} + 16)^{\frac{3}{2}} - 48 x^{2} + 16

Add a constant to the solution

= (x^{2} + 16)^{\frac{3}{2}} - 48 x^{2} + 16 + C

\frac{d}{d x} (\frac{e ^{x}}{cos ( 3 x )})

\int cot^{2} (x) csc^{2} (x) d x

\int 7 x^{3} e^{x^{4}} d x

\frac{d}{d x} ((4 x^{2} - 5)^{3})

x y^{^{'}} - 2 y - 8 = 0

What is the integral of (3x^3)/(sqrt(x^2+16)) ?
The integral of (3x^3)/(sqrt(x^2+16)) is (x^2+16)^{3/2}-48sqrt(x^2+16)+C