What is the integral of (x^2+4x+10)/(x^3+2x^2+5x) ?

The integral of (x^2+4x+10)/(x^3+2x^2+5x) is 2ln|x|-1/2 ln|x^2+2x+5|+1/2 arctan(1/2 (x+1))+C

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integral of (x^2+4x+10)/(x^3+2x^2+5x)

\int \frac{x ^{2} + 4 x + 10}{x ^{3} + 2 x ^{2} + 5 x} d x

2 ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 2 x + 5 + \frac{1}{2} arctan (\frac{1}{2} (x + 1)) + C

Solution steps

\int \frac{x ^{2} + 4 x + 10}{x ^{3} + 2 x ^{2} + 5 x} d x

Take the partial fraction of

\frac{x ^{2} + 4 x + 10}{x ^{3} + 2 x ^{2} + 5 x} : \frac{2}{x} - \frac{x}{x ^{2} + 2 x + 5}

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

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

\int \frac{x}{x ^{2} + 2 x + 5} d x = \frac{1}{2} ln x^{2} + 2 x + 5 - \frac{1}{2} arctan (\frac{1}{2} (x + 1))

= 2 ln ∣ x ∣ - (\frac{1}{2} ln x^{2} + 2 x + 5 - \frac{1}{2} arctan (\frac{1}{2} (x + 1)))

Simplify

= 2 ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 2 x + 5 + \frac{1}{2} arctan (\frac{1}{2} (x + 1))

Add a constant to the solution

= 2 ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 2 x + 5 + \frac{1}{2} arctan (\frac{1}{2} (x + 1)) + C

\frac{d}{d x} (ln (x (1 - x)))

ln (X^{\frac{1}{2}})

\int \frac{6}{( x + 1 ) ^{2}} d x

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

\frac{s ^{2} - 4 s + 4}{s ^{2} - 4}

What is the integral of (x^2+4x+10)/(x^3+2x^2+5x) ?
The integral of (x^2+4x+10)/(x^3+2x^2+5x) is 2ln|x|-1/2 ln|x^2+2x+5|+1/2 arctan(1/2 (x+1))+C