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

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

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

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

x + 2 - 2 ln x^{2} + 4 x + 5 + 3 arctan (x + 2) + C

Solution steps

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

Long division

\frac{x ^{2}}{x ^{2} + 4 x + 5} : 1 + \frac{- 4 x - 5}{x ^{2} + 4 x + 5}

= \int 1 + \frac{- 4 x - 5}{x ^{2} + 4 x + 5} d x

Complete the square

x^{2} + 4 x + 5 : (x + 2)^{2} + 1

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

Apply u-substitution

: \int 1 + \frac{- 4 u + 3}{u ^{2} + 1} d u

= \int 1 + \frac{- 4 u + 3}{u ^{2} + 1} d u

Apply the Sum Rule:

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

= \int 1 d u + \int \frac{- 4 u + 3}{u ^{2} + 1} d u

\int 1 d u = u

\int \frac{- 4 u + 3}{u ^{2} + 1} d u = - 2 ln u^{2} + 1 + 3 arctan (u)

= u - 2 ln u^{2} + 1 + 3 arctan (u)

Substitute back

u = x + 2

= x + 2 - 2 ln (x + 2)^{2} + 1 + 3 arctan (x + 2)

Expand

(x + 2)^{2} + 1 : x^{2} + 4 x + 5

= x + 2 - 2 ln x^{2} + 4 x + 5 + 3 arctan (x + 2)

Add a constant to the solution

= x + 2 - 2 ln x^{2} + 4 x + 5 + 3 arctan (x + 2) + C

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