What is the integral of 1/(2x^2-4x+12) ?

The integral of 1/(2x^2-4x+12) is 1/(2sqrt(5))arctan(1/(sqrt(5))(x-1))+C

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integral of 1/(2x^2-4x+12)

\int \frac{1}{2 x ^{2} - 4 x + 12} d x

\frac{1}{2 5} arctan (\frac{1}{5} (x - 1)) + C

Solution steps

\int \frac{1}{2 x ^{2} - 4 x + 12} d x

Complete the square

2 x^{2} - 4 x + 12 : 2 (x - 1)^{2} + 10

= \int \frac{1}{2 ( x - 1 ) ^{2} + 10} d x

Apply u-substitution

: \int \frac{1}{2 u ^{2} + 10} d u

= \int \frac{1}{2 u ^{2} + 10} d u

Apply Integral Substitution

= \int \frac{1}{2 5 ( v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{2 5} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= \frac{1}{2 5} arctan (v)

Substitute back

= \frac{1}{2 5} arctan (\frac{2}{10} (x - 1))

Simplify

\frac{1}{2 5} arctan (\frac{2}{10} (x - 1)) : \frac{1}{2 5} arctan (\frac{1}{5} (x - 1))

= \frac{1}{2 5} arctan (\frac{1}{5} (x - 1))

Add a constant to the solution

= \frac{1}{2 5} arctan (\frac{1}{5} (x - 1)) + C

\frac{d}{d x} (x^{3} + y^{2})

\int y e^{- y^{2}} d y

y = - 3 - 7 x^{2}

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

\frac{\partial}{\partial x} (2 e^{x^{2}})

What is the integral of 1/(2x^2-4x+12) ?
The integral of 1/(2x^2-4x+12) is 1/(2sqrt(5))arctan(1/(sqrt(5))(x-1))+C