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

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

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

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

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

Solution steps

\int x^{3} 1 + x^{4} 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 = 1 + x^{4}

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

Simplify

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

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

Add a constant to the solution

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

f (x) = \frac{1}{x - 1}, at x = 4

\int 4 x sin (2 x^{2}) d x

\frac{\partial}{\partial x} (x^{6} + 8 x y^{5})

\frac{\partial}{\partial x} (3 x + 4 y)

(25 t) (\frac{t ^{2}}{25})

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