What is the integral of (x^2)/(x^3+8) ?

The integral of (x^2)/(x^3+8) is 1/3 ln|x^3+8|+C

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integral of (x^2)/(x^3+8)

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

\frac{1}{3} ln x^{3} + 8 + C

Solution steps

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

Apply u-substitution

= \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} + 8

= \frac{1}{3} ln x^{3} + 8

Add a constant to the solution

= \frac{1}{3} ln x^{3} + 8 + C

x \to 1 lim (\frac{x + 3}{x - 1})

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

\int \frac{26}{26 + e ^{x}} d x

t an g e n t \frac{x}{x + 6}

\int_{6}^{8} \frac{24}{( x - 6 ) ^{3}} d x

What is the integral of (x^2)/(x^3+8) ?
The integral of (x^2)/(x^3+8) is 1/3 ln|x^3+8|+C