What is the integral of (x^3)/(x^4-5) ?

The integral of (x^3)/(x^4-5) is 1/4 ln|x^4-5|+C

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integral of (x^3)/(x^4-5)

\int \frac{x ^{3}}{x ^{4} - 5} d x

\frac{1}{4} ln x^{4} - 5 + C

Solution steps

\int \frac{x ^{3}}{x ^{4} - 5} 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} - 5

= \frac{1}{4} ln x^{4} - 5

Add a constant to the solution

= \frac{1}{4} ln x^{4} - 5 + C

\int \frac{x ^{2} - 18 x + 32}{x ( x - 4 ) ^{2}} d x

f (x) = x^{\frac{2}{5}}

2 sin (3 t) + 5 cos (3 t)

y^{^{''}} + 13 y = 0

\frac{d y}{d x} = \frac{x + x y ^{2}}{4 y}

What is the integral of (x^3)/(x^4-5) ?
The integral of (x^3)/(x^4-5) is 1/4 ln|x^4-5|+C