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

The integral of (2x^3)/(sqrt(x^2-4)) is 2(1/3 (x^2-4)^{3/2}+4sqrt(x^2-4))+C

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integral of (2x^3)/(sqrt(x^2-4))

\int \frac{2 x ^{3}}{x ^{2} - 4} d x

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

Solution steps

\int \frac{2 x ^{3}}{x ^{2} - 4} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{x ^{3}}{x ^{2} - 4} d x

Apply u-substitution

= 2 \cdot \int u^{2} + 4 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

= 2 (\int u^{2} d u + \int 4 d u)

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

\int 4 d u = 4 u

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

Substitute back

u = x^{2} - 4

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

Simplify

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

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

Add a constant to the solution

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

What is the integral of (2x^3)/(sqrt(x^2-4)) ?
The integral of (2x^3)/(sqrt(x^2-4)) is 2(1/3 (x^2-4)^{3/2}+4sqrt(x^2-4))+C