What is the integral of (x^3)/(sqrt(x^4+12)) ?

The integral of (x^3)/(sqrt(x^4+12)) is 1/2 sqrt(x^4+12)+C

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integral of (x^3)/(sqrt(x^4+12))

\int \frac{x ^{3}}{x ^{4} + 12} d x

\frac{1}{2} x^{4} + 12 + C

Solution steps

\int \frac{x ^{3}}{x ^{4} + 12} d x

Apply u-substitution

: \int \frac{1}{4 u} d u

= \int \frac{1}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= \frac{1}{4} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = x^{4} + 12

= \frac{1}{4} \cdot 2 (x^{4} + 12)^{\frac{1}{2}}

Simplify

\frac{1}{4} \cdot 2 (x^{4} + 12)^{\frac{1}{2}} : \frac{1}{2} x^{4} + 12

= \frac{1}{2} x^{4} + 12

Add a constant to the solution

= \frac{1}{2} x^{4} + 12 + C

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What is the integral of (x^3)/(sqrt(x^4+12)) ?
The integral of (x^3)/(sqrt(x^4+12)) is 1/2 sqrt(x^4+12)+C