What is the integral of (2x^3+8x)/(sqrt(x^2+4)) ?

The integral of (2x^3+8x)/(sqrt(x^2+4)) is 2/3 (x^2+4)^{3/2}+C

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integral of (2x^3+8x)/(sqrt(x^2+4))

\int \frac{2 x ^{3} + 8 x}{x ^{2} + 4} d x

\frac{2}{3} (x^{2} + 4)^{\frac{3}{2}} + C

Solution steps

\int \frac{2 x ^{3} + 8 x}{x ^{2} + 4} d x

Expand

\frac{2 x ^{3} + 8 x}{x ^{2} + 4} : \frac{2 x ^{3}}{x ^{2} + 4} + \frac{8 x}{x ^{2} + 4}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{2 x ^{3}}{x ^{2} + 4} d x + \int \frac{8 x}{x ^{2} + 4} d x

\int \frac{2 x ^{3}}{x ^{2} + 4} d x = 2 (\frac{1}{3} (x^{2} + 4)^{\frac{3}{2}} - 4 x^{2} + 4)

\int \frac{8 x}{x ^{2} + 4} d x = 8 x^{2} + 4

= 2 (\frac{1}{3} (x^{2} + 4)^{\frac{3}{2}} - 4 x^{2} + 4) + 8 x^{2} + 4

Simplify

2 (\frac{1}{3} (x^{2} + 4)^{\frac{3}{2}} - 4 x^{2} + 4) + 8 x^{2} + 4 : \frac{2}{3} (x^{2} + 4)^{\frac{3}{2}}

= \frac{2}{3} (x^{2} + 4)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{3} (x^{2} + 4)^{\frac{3}{2}} + C

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What is the integral of (2x^3+8x)/(sqrt(x^2+4)) ?
The integral of (2x^3+8x)/(sqrt(x^2+4)) is 2/3 (x^2+4)^{3/2}+C