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

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

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

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

\frac{1}{3} ln x^{3} + 3 x^{2} - 5 + C

Solution steps

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

= \frac{1}{3} ln x^{3} + 3 x^{2} - 5

Add a constant to the solution

= \frac{1}{3} ln x^{3} + 3 x^{2} - 5 + C

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\int \frac{1}{sin ^{4} ( x ) cos ^{2} ( x )} d x

\frac{\partial}{\partial x} (x^{6} + 2^{y} + x^{y})

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

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

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