What is the integral of (x^2)/(x^3+4x^2+6x) ?

The integral of (x^2)/(x^3+4x^2+6x) is 1/2 ln| 1/2 (x+2)^2+1|-sqrt(2)arctan(1/(sqrt(2))(x+2))+C

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integral of (x^2)/(x^3+4x^2+6x)

\int \frac{x ^{2}}{x ^{3} + 4 x ^{2} + 6 x} d x

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

Solution steps

\int \frac{x ^{2}}{x ^{3} + 4 x ^{2} + 6 x} d x

Factor

\frac{x ^{2}}{x ^{3} + 4 x ^{2} + 6 x}

= \int \frac{x ^{2}}{x ( x ^{2} + 4 x + 6 )} d x

Cancel the common factor:

x

= \int \frac{x}{x ^{2} + 4 x + 6} d x

Complete the square

x^{2} + 4 x + 6 : (x + 2)^{2} + 2

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

Apply u-substitution

= \int \frac{u - 2}{u ^{2} + 2} d u

Apply Integral Substitution

= \int \frac{v - 2}{v ^{2} + 1} d v

Expand

\frac{v - 2}{v ^{2} + 1} : \frac{v}{v ^{2} + 1} - \frac{2}{v ^{2} + 1}

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

\int \frac{v}{v ^{2} + 1} d v = \frac{1}{2} ln v^{2} + 1

\int \frac{2}{v ^{2} + 1} d v = 2 arctan (v)

= \frac{1}{2} ln v^{2} + 1 - 2 arctan (v)

Substitute back

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

Simplify

\frac{1}{2} ln (\frac{x + 2}{2})^{2} + 1 - 2 arctan (\frac{x + 2}{2}) : \frac{1}{2} ln \frac{1}{2} (x + 2)^{2} + 1 - 2 arctan (\frac{1}{2} (x + 2))

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

Add a constant to the solution

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

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What is the integral of (x^2)/(x^3+4x^2+6x) ?
The integral of (x^2)/(x^3+4x^2+6x) is 1/2 ln| 1/2 (x+2)^2+1|-sqrt(2)arctan(1/(sqrt(2))(x+2))+C