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

The integral of x^3sqrt(x^4+2) is 1/6 (x^4+2)^{3/2}+C

Popular >

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

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

\frac{1}{6} (x^{4} + 2)^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

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

Substitute back

u = x^{4} + 2

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

Simplify

\frac{1}{4} \cdot \frac{2}{3} (x^{4} + 2)^{\frac{3}{2}} : \frac{1}{6} (x^{4} + 2)^{\frac{3}{2}}

= \frac{1}{6} (x^{4} + 2)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{6} (x^{4} + 2)^{\frac{3}{2}} + C

(3 - 5 x e^{x}) (3 x + 2)

\frac{\partial}{\partial x} (e^{- \frac{x ^{2}}{4 d ^{2}}})

(1 + cos (t)) (\frac{d y}{d t}) = (1 + e^{- y}) sin (t)

f (x) = (x^{2} + 36) (2 x + 21)

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

What is the integral of x^3sqrt(x^4+2) ?
The integral of x^3sqrt(x^4+2) is 1/6 (x^4+2)^{3/2}+C