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

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

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

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

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

Solution steps

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

Apply radical rule:

a = a^{\frac{1}{2}}

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

Apply u-substitution

: \int \frac{u ^{\frac{1}{3}}}{3 u ^{\frac{4}{3}} - 4} d u

= \int \frac{u ^{\frac{1}{3}}}{3 u ^{\frac{4}{3}} - 4} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{u ^{\frac{1}{3}}}{u ^{\frac{4}{3}} - 4} d u

Apply u-substitution

: \int \frac{3}{4 v} d v

= \frac{1}{3} \cdot \int \frac{3}{4 v} d v

Take the constant out:

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

= \frac{1}{3} \cdot \frac{3}{4} \cdot \int \frac{1}{v} d v

Apply the Power Rule

: 2 v^{\frac{1}{2}}

= \frac{1}{3} \cdot \frac{3}{4} \cdot 2 v^{\frac{1}{2}}

Substitute back

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

Simplify

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

= \frac{1}{2} x^{4} - 4

Add a constant to the solution

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

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