What is the integral of 1/(au^2+bu+c) ?

The integral of 1/(au^2+bu+c) is 2/(sqrt(-b^2+4ac))arctan((2au+b)/(sqrt(4ac-b^2)))+C

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integral of 1/(au^2+bu+c)

\int \frac{1}{a u ^{2} + b u + c} d u

\frac{2}{- b ^{2} + 4 a c} arctan (\frac{2 a u + b}{4 a c - b ^{2}}) + C

Solution steps

\int \frac{1}{a u ^{2} + b u + c} d u

Complete the square

a u^{2} + b u + c : (a u + \frac{b}{2 a})^{2} + c - \frac{b ^{2}}{4 a}

= \int \frac{1}{( a u + \frac{b}{2 a} ) ^{2} + c - \frac{b ^{2}}{4 a}} d u

Apply u-substitution

= \int \frac{4 a}{4 a v ^{2} + 4 a c - b ^{2}} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 a \cdot \int \frac{1}{4 a v ^{2} + 4 a c - b ^{2}} d v

Apply Integral Substitution

= 4 a \cdot \int \frac{1}{2 a ( w ^{2} + 1 ) - b ^{2} + 4 a c} d w

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 a \frac{1}{2 a - b ^{2} + 4 a c} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

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

= 4 a \frac{1}{2 a - b ^{2} + 4 a c} arctan (w)

Substitute back

= 4 a \frac{1}{2 a - b ^{2} + 4 a c} arctan (\frac{2 a}{4 a c - b ^{2}} (a u + \frac{b}{2 a}))

Simplify

4 a \frac{1}{2 a - b ^{2} + 4 a c} arctan (\frac{2 a}{4 a c - b ^{2}} (a u + \frac{b}{2 a})) : \frac{2}{- b ^{2} + 4 a c} arctan (\frac{2 a u + b}{4 a c - b ^{2}})

= \frac{2}{- b ^{2} + 4 a c} arctan (\frac{2 a u + b}{4 a c - b ^{2}})

Add a constant to the solution

= \frac{2}{- b ^{2} + 4 a c} arctan (\frac{2 a u + b}{4 a c - b ^{2}}) + C

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What is the integral of 1/(au^2+bu+c) ?
The integral of 1/(au^2+bu+c) is 2/(sqrt(-b^2+4ac))arctan((2au+b)/(sqrt(4ac-b^2)))+C