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

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

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integral of (x^3)/(sqrt(2x^4+1))

\int \frac{x ^{3}}{2 x ^{4} + 1} d x

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

Solution steps

\int \frac{x ^{3}}{2 x ^{4} + 1} d x

Apply u-substitution

: \int \frac{1}{8 u} d u

= \int \frac{1}{8 u} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= \frac{1}{8} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 2 x^{4} + 1

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

Simplify

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

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

Add a constant to the solution

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

\int (2 x - 5) d x

ln (x^{3} + 1)

\int 0.5 x^{2} d x

x \to \infty lim (- 1^{x})

\frac{d}{d x} (arccos (e^{x}))

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