What is the integral of 1/(ax^2+bx+c) ?

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

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integral of 1/(ax^2+bx+c)

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int \frac{4 a}{4 a u ^{2} + 4 a c - b ^{2}} d u

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

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 u ^{2} + 4 a c - b ^{2}} d u

Apply Integral Substitution

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

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}{v ^{2} + 1} d v

Use the common integral:

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

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

Substitute back

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

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

Add a constant to the solution

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

y^{^{'}} - (\frac{2 x}{1 + x ^{2}}) y = 0, y (0) = 4

\frac{d}{d x} (\frac{x ^{3}}{3} - x^{2} - 3 x)

(t^{2} - t) (t^{2} + t)

\frac{\partial}{\partial y} (e^{3 x y})

f (x) = x - 4, (20, 40)

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