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

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

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

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

n = 1 \sum \infty \frac{( x ^{n} )}{n !}

sin^{2} (1 - s^{2})

\int tan^{2} (θ) sec^{4} (θ) d θ

x^{3} - 3 x, x

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