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

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

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

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (1 - cos (2 x) + 2 cos^{2} (x))

\int 6 x^{13} e^{- x^{7}} d x

x \to 0 lim (\frac{4}{1 + 2 ^{\frac{1}{x}}})

\frac{d}{d x} (4 cos (4 x))

f^{^{'}} (x) = 3 x - 4, f (- 1) = \frac{13}{2}

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