What is the integral of 1/(2x^2-4x+20) ?

The integral of 1/(2x^2-4x+20) is 1/6 arctan((x-1)/3)+C

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integral of 1/(2x^2-4x+20)

\int \frac{1}{2 x ^{2} - 4 x + 20} d x

\frac{1}{6} arctan (\frac{x - 1}{3}) + C

Solution steps

\int \frac{1}{2 x ^{2} - 4 x + 20} d x

Complete the square

2 x^{2} - 4 x + 20 : 2 (x - 1)^{2} + 18

= \int \frac{1}{2 ( x - 1 ) ^{2} + 18} d x

Apply u-substitution

: \int \frac{1}{2 u ^{2} + 18} d u

= \int \frac{1}{2 u ^{2} + 18} d u

Apply Integral Substitution

= \int \frac{1}{6 ( 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}{6} \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}{6} arctan (v)

Substitute back

= \frac{1}{6} arctan (\frac{2}{3 2} (x - 1))

Simplify

\frac{1}{6} arctan (\frac{2}{3 2} (x - 1)) : \frac{1}{6} arctan (\frac{x - 1}{3})

= \frac{1}{6} arctan (\frac{x - 1}{3})

Add a constant to the solution

= \frac{1}{6} arctan (\frac{x - 1}{3}) + C

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

x \to - 3.1 lim (x^{2})

\frac{d y}{d x} = \frac{x x ^{2} - y ^{2} + y ^{2}}{x y}

\int sin (\frac{x y}{2}) d y

x \to 0 lim (\frac{cos ( x )}{5 x})

What is the integral of 1/(2x^2-4x+20) ?
The integral of 1/(2x^2-4x+20) is 1/6 arctan((x-1)/3)+C