What is the integral of (x^3)/(sqrt(x^2+8)) ?

The integral of (x^3)/(sqrt(x^2+8)) is 1/3 (x^2+8)^{3/2}-8sqrt(x^2+8)+C

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integral of (x^3)/(sqrt(x^2+8))

\int \frac{x ^{3}}{x ^{2} + 8} d x

\frac{1}{3} (x^{2} + 8)^{\frac{3}{2}} - 8 x^{2} + 8 + C

Solution steps

\int \frac{x ^{3}}{x ^{2} + 8} d x

Apply u-substitution

= \int u^{2} - 8 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 8 d u

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

\int 8 d u = 8 u

= \frac{u ^{3}}{3} - 8 u

Substitute back

u = x^{2} + 8

= \frac{( x ^{2} + 8 ) ^{3}}{3} - 8 x^{2} + 8

Simplify

\frac{( x ^{2} + 8 ) ^{3}}{3} - 8 x^{2} + 8 : \frac{1}{3} (x^{2} + 8)^{\frac{3}{2}} - 8 x^{2} + 8

= \frac{1}{3} (x^{2} + 8)^{\frac{3}{2}} - 8 x^{2} + 8

Add a constant to the solution

= \frac{1}{3} (x^{2} + 8)^{\frac{3}{2}} - 8 x^{2} + 8 + C

What is the integral of (x^3)/(sqrt(x^2+8)) ?
The integral of (x^3)/(sqrt(x^2+8)) is 1/3 (x^2+8)^{3/2}-8sqrt(x^2+8)+C