What is the integral of (5x^2)/(x^2+x-6) ?

The integral of (5x^2)/(x^2+x-6) is 5x+4ln|x-2|-9ln|x+3|+C

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integral of (5x^2)/(x^2+x-6)

\int \frac{5 x ^{2}}{x ^{2} + x - 6} d x

5 x + 4 ln ∣ x - 2 ∣ - 9 ln ∣ x + 3 ∣ + C

Solution steps

\int \frac{5 x ^{2}}{x ^{2} + x - 6} d x

Take the constant out:

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

= 5 \cdot \int \frac{x ^{2}}{x ^{2} + x - 6} d x

Take the partial fraction of

\frac{x ^{2}}{x ^{2} + x - 6} : 1 + \frac{4}{5 ( x - 2 )} - \frac{9}{5 ( x + 3 )}

= 5 \cdot \int 1 + \frac{4}{5 ( x - 2 )} - \frac{9}{5 ( x + 3 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 5 (\int 1 d x + \int \frac{4}{5 ( x - 2 )} d x - \int \frac{9}{5 ( x + 3 )} d x)

\int 1 d x = x

\int \frac{4}{5 ( x - 2 )} d x = \frac{4}{5} ln ∣ x - 2 ∣

\int \frac{9}{5 ( x + 3 )} d x = \frac{9}{5} ln ∣ x + 3 ∣

= 5 (x + \frac{4}{5} ln ∣ x - 2 ∣ - \frac{9}{5} ln ∣ x + 3 ∣)

Simplify

5 (x + \frac{4}{5} ln ∣ x - 2 ∣ - \frac{9}{5} ln ∣ x + 3 ∣) : 5 x + 4 ln ∣ x - 2 ∣ - 9 ln ∣ x + 3 ∣

= 5 x + 4 ln ∣ x - 2 ∣ - 9 ln ∣ x + 3 ∣

Add a constant to the solution

= 5 x + 4 ln ∣ x - 2 ∣ - 9 ln ∣ x + 3 ∣ + C

\frac{\partial}{\partial x} (2 x cos (2 y) - y^{2})

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

x \to 0 lim (\frac{g ( x )}{x})

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

\int_{1}^{2} (x^{3} + \frac{1}{4 x ^{3}}) d x

What is the integral of (5x^2)/(x^2+x-6) ?
The integral of (5x^2)/(x^2+x-6) is 5x+4ln|x-2|-9ln|x+3|+C