What is the integral of x^3sqrt(9x^4+1) ?

The integral of x^3sqrt(9x^4+1) is 1/54 (9x^4+1)^{3/2}+C

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integral of x^3sqrt(9x^4+1)

\int x^{3} 9 x^{4} + 1 d x

\frac{1}{54} (9 x^{4} + 1)^{\frac{3}{2}} + C

Solution steps

\int x^{3} 9 x^{4} + 1 d x

Apply u-substitution

: \int \frac{u}{36} d u

= \int \frac{u}{36} d u

Take the constant out:

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

= \frac{1}{36} \cdot \int u d u

Apply the Power Rule

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

= \frac{1}{36} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 9 x^{4} + 1

= \frac{1}{36} \cdot \frac{2}{3} (9 x^{4} + 1)^{\frac{3}{2}}

Simplify

\frac{1}{36} \cdot \frac{2}{3} (9 x^{4} + 1)^{\frac{3}{2}} : \frac{1}{54} (9 x^{4} + 1)^{\frac{3}{2}}

= \frac{1}{54} (9 x^{4} + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{54} (9 x^{4} + 1)^{\frac{3}{2}} + C

\int 6 x sec^{2} (x) d x

x y + 2 x \frac{d y}{d x} = y

x \to 0 + lim (\frac{1}{x} - ln (x))

n = 0 \sum \infty (\frac{x + 9}{2})^{n}

x \to 0 lim (\frac{x ^{2}}{3 x})

What is the integral of x^3sqrt(9x^4+1) ?
The integral of x^3sqrt(9x^4+1) is 1/54 (9x^4+1)^{3/2}+C