What is the integral of (x^3)/(x^4+10) ?

The integral of (x^3)/(x^4+10) is 1/4 ln|x^4+10|+C

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integral of (x^3)/(x^4+10)

\int \frac{x ^{3}}{x ^{4} + 10} d x

\frac{1}{4} ln x^{4} + 10 + C

Solution steps

\int \frac{x ^{3}}{x ^{4} + 10} d x

Apply u-substitution

: \int \frac{1}{4 u} d u

= \int \frac{1}{4 u} 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 \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} ln ∣ u ∣

Substitute back

u = x^{4} + 10

= \frac{1}{4} ln x^{4} + 10

Add a constant to the solution

= \frac{1}{4} ln x^{4} + 10 + C

(sin^{3} (t))^{^{'}}

y^{^{''}} = - 4 y + e^{x}

\int \frac{x ^{2} - 3 x - 8}{x ^{2} - 2 x + 1} d x

x y - 6 y^{2} = 4, (14, 2)

\int x^{n - 1} a + b x^{n} d x

What is the integral of (x^3)/(x^4+10) ?
The integral of (x^3)/(x^4+10) is 1/4 ln|x^4+10|+C