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

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

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

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d ^{2} y}{d x ^{2}} + a^{2} \cdot y - e^{b - x} - 1 = 0

\frac{1}{y} d x - \frac{x}{( ln ( x ) ) ^{3}} d y = 0

\frac{1}{y ^{2}} - \frac{3}{y ^{4}} (y + 5 y^{3})

\int 3 x^{- 3} + sin (x) d x

\frac{d y}{d x} = \frac{y}{x} + \frac{x}{y}

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