What is the integral of 1/(2x^2-4x+14) ?

The integral of 1/(2x^2-4x+14) is 1/(2sqrt(6))arctan(1/(sqrt(6))(x-1))+C

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integral of 1/(2x^2-4x+14)

\int \frac{1}{2 x ^{2} - 4 x + 14} d x

\frac{1}{2 6} arctan (\frac{1}{6} (x - 1)) + C

Solution steps

\int \frac{1}{2 x ^{2} - 4 x + 14} d x

Complete the square

2 x^{2} - 4 x + 14 : 2 (x - 1)^{2} + 12

= \int \frac{1}{2 ( x - 1 ) ^{2} + 12} d x

Apply u-substitution

: \int \frac{1}{2 u ^{2} + 12} d u

= \int \frac{1}{2 u ^{2} + 12} d u

Apply Integral Substitution

= \int \frac{1}{2 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}{2 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}{2 6} arctan (v)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int x 5^{x} d x

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

\int_{0}^{1} \frac{8 + x}{1 + x ^{2}} d x

\frac{5 + s}{s ( s ^{2} - 5 s + 4 )}

\frac{d}{d x} (\frac{3}{5})

What is the integral of 1/(2x^2-4x+14) ?
The integral of 1/(2x^2-4x+14) is 1/(2sqrt(6))arctan(1/(sqrt(6))(x-1))+C