What is the integral of (x^2)/(x^3-5) ?

The integral of (x^2)/(x^3-5) is 1/3 ln|x^3-5|+C

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integral of (x^2)/(x^3-5)

\int \frac{x ^{2}}{x ^{3} - 5} d x

\frac{1}{3} ln x^{3} - 5 + C

Solution steps

\int \frac{x ^{2}}{x ^{3} - 5} d x

Apply u-substitution

: \int \frac{1}{3 u} d u

= \int \frac{1}{3 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = x^{3} - 5

= \frac{1}{3} ln x^{3} - 5

Add a constant to the solution

= \frac{1}{3} ln x^{3} - 5 + C

\frac{1}{2 x}, at 2

x \to 0 lim (\frac{( sin ( 2 x ) )}{x})

\frac{d y}{d x} = - y x^{2}

\frac{d}{d x} (- (x^{2} + 3 x - 2)^{3})

y = x sin (x) + cot (x)

What is the integral of (x^2)/(x^3-5) ?
The integral of (x^2)/(x^3-5) is 1/3 ln|x^3-5|+C