What is the integral of (x^2)/(x^3-7) ?

The integral of (x^2)/(x^3-7) is 1/3 ln|x^3-7|+C

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integral of (x^2)/(x^3-7)

\int \frac{x ^{2}}{x ^{3} - 7} d x

\frac{1}{3} ln x^{3} - 7 + C

Solution steps

\int \frac{x ^{2}}{x ^{3} - 7} 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} - 7

= \frac{1}{3} ln x^{3} - 7

Add a constant to the solution

= \frac{1}{3} ln x^{3} - 7 + C

\int_{0}^{4} \frac{1}{x} d x

\frac{3}{8 x ^{\frac{5}{2}}}

\int (t^{2} - 2 t^{4}) d t

\int_{1}^{2} \frac{1}{x ^{5}} d x

\frac{\partial}{\partial x} (y e^{4 x})

What is the integral of (x^2)/(x^3-7) ?
The integral of (x^2)/(x^3-7) is 1/3 ln|x^3-7|+C