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

The integral of (x^3)/(1+9x^4) is 1/36 ln|1+9x^4|+C

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

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

\frac{1}{36} ln 1 + 9 x^{4} + C

Solution steps

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

Apply u-substitution

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{36} ln ∣ u ∣

Substitute back

u = 1 + 9 x^{4}

= \frac{1}{36} ln 1 + 9 x^{4}

Add a constant to the solution

= \frac{1}{36} ln 1 + 9 x^{4} + C

x \to 4 lim (\frac{2 x - 8}{x - 4})

\int x^{2} \cdot (x - 5) \cdot 7 - x d x

\int_{0}^{10} ∣ x - 5 ∣ d x

\frac{1}{2 x + 2}

\int x e^{9 x^{2}} d x

What is the integral of (x^3)/(1+9x^4) ?
The integral of (x^3)/(1+9x^4) is 1/36 ln|1+9x^4|+C