What is the integral of x/(\sqrt[4]{x^2+2)} ?

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

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integral of x/(\sqrt[4]{x^2+2)}

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

Solution steps

Apply u-substitution

Take the constant out:

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

Apply u-substitution

Apply the Power Rule

: \frac{4}{3} v^{\frac{3}{4}}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

2 (x - 6)^{3}

z \to 0 lim (z^{3} cos (\frac{1}{z}))

f (x) = 3 sec (x), at x = \frac{π}{3}

cos^{3} (e^{4 x})

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