What is the integral of ((x-1)^2)/(x^2+3x+4) ?

The integral of ((x-1)^2)/(x^2+3x+4) is x-5/2 ln|4x^2+12x+16|+9/(sqrt(7))arctan(1/(sqrt(7))(2x+3))+C

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integral of ((x-1)^2)/(x^2+3x+4)

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

x - \frac{5}{2} ln 4 x^{2} + 12 x + 16 + \frac{9}{7} arctan (\frac{1}{7} (2 x + 3)) + C

Solution steps

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

Long division

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

= \int 1 + \frac{- 5 x - 3}{x ^{2} + 3 x + 4} 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 1 d x + \int \frac{- 5 x - 3}{x ^{2} + 3 x + 4} d x

\int 1 d x = x

\int \frac{- 5 x - 3}{x ^{2} + 3 x + 4} d x = - \frac{5}{2} ln 4 x^{2} + 12 x + 16 + \frac{9}{7} arctan (\frac{1}{7} (2 x + 3))

= x - \frac{5}{2} ln 4 x^{2} + 12 x + 16 + \frac{9}{7} arctan (\frac{1}{7} (2 x + 3))

Add a constant to the solution

= x - \frac{5}{2} ln 4 x^{2} + 12 x + 16 + \frac{9}{7} arctan (\frac{1}{7} (2 x + 3)) + C

\int \frac{4 t ^{3} - t ^{2} + 36 t}{t ^{2} + 9} d t

\frac{\partial}{\partial x} (z - x y)

\frac{\partial}{\partial x} (sin (5 x + 6 y + z))

\int sin (5 x) cos (2 x) d x

x \to 9 + lim (\frac{3}{x - 9})

What is the integral of ((x-1)^2)/(x^2+3x+4) ?
The integral of ((x-1)^2)/(x^2+3x+4) is x-5/2 ln|4x^2+12x+16|+9/(sqrt(7))arctan(1/(sqrt(7))(2x+3))+C