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

\int_{1}^{2} \frac{12}{x ^{2}} - 1 d x

\frac{d}{d x} (x^{2} y z)

\frac{d}{d x} (ln (4^{x}))

\int \frac{6}{x ^{2} x ^{2} + 9} d x

d x = e^{3 x} d y

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