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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \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

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

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

\int 16 d u = 16 u

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

Substitute back

u = 16 + x^{2}

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

Simplify

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

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

Add a constant to the solution

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

x \to - 2 lim ((x + 1)^{4} (2 x^{2}))

\frac{x ^{2} + 9}{x + 3}

\int_{0}^{2} \frac{1}{4 + x ^{2}} d x

f (x) = \frac{2}{x + 1}, at x = 1

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