What is the integral of (6x^3)/(sqrt(16+x^2)) ?

The integral of (6x^3)/(sqrt(16+x^2)) is 6(1/3 (16+x^2)^{3/2}-16sqrt(16+x^2))+C

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integral of (6x^3)/(sqrt(16+x^2))

\int \frac{6 x ^{3}}{16 + x ^{2}} d x

6 (\frac{1}{3} (16 + x^{2})^{\frac{3}{2}} - 16 16 + x^{2}) + C

Solution steps

\int \frac{6 x ^{3}}{16 + x ^{2}} 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}}{16 + x ^{2}} d x

Apply u-substitution

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

= 6 \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

= 6 (\int u^{2} d u - \int 16 d u)

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

\int 16 d u = 16 u

= 6 (\frac{u ^{3}}{3} - 16 u)

Substitute back

u = 16 + x^{2}

= 6 (\frac{( 16 + x ^{2} ) ^{3}}{3} - 16 16 + x^{2})

Simplify

6 (\frac{( 16 + x ^{2} ) ^{3}}{3} - 16 16 + x^{2}) : 6 (\frac{1}{3} (16 + x^{2})^{\frac{3}{2}} - 16 16 + x^{2})

= 6 (\frac{1}{3} (16 + x^{2})^{\frac{3}{2}} - 16 16 + x^{2})

Add a constant to the solution

= 6 (\frac{1}{3} (16 + x^{2})^{\frac{3}{2}} - 16 16 + x^{2}) + C

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

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\int - 6 ln (x^{2} - 1) d x

\int (x cos (4 x)) d x

What is the integral of (6x^3)/(sqrt(16+x^2)) ?
The integral of (6x^3)/(sqrt(16+x^2)) is 6(1/3 (16+x^2)^{3/2}-16sqrt(16+x^2))+C