What is the integral of 1/(2x^2-12x+26) ?

The integral of 1/(2x^2-12x+26) is 1/4 arctan((x-3)/2)+C

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integral of 1/(2x^2-12x+26)

\int \frac{1}{2 x ^{2} - 12 x + 26} d x

\frac{1}{4} arctan (\frac{x - 3}{2}) + C

Solution steps

\int \frac{1}{2 x ^{2} - 12 x + 26} d x

Complete the square

2 x^{2} - 12 x + 26 : 2 (x - 3)^{2} + 8

= \int \frac{1}{2 ( x - 3 ) ^{2} + 8} d x

Apply u-substitution

: \int \frac{1}{2 u ^{2} + 8} d u

= \int \frac{1}{2 u ^{2} + 8} d u

Apply Integral Substitution

= \int \frac{1}{4 ( 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}{4} \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}{4} arctan (v)

Substitute back

= \frac{1}{4} arctan (\frac{2}{2 2} (x - 3))

Simplify

\frac{1}{4} arctan (\frac{2}{2 2} (x - 3)) : \frac{1}{4} arctan (\frac{x - 3}{2})

= \frac{1}{4} arctan (\frac{x - 3}{2})

Add a constant to the solution

= \frac{1}{4} arctan (\frac{x - 3}{2}) + C

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

\frac{\partial}{\partial y} (\frac{1}{2} (x - y)^{2})

x \to 2 lim (x^{2} - 4 x + 2)

\frac{1}{2 s ^{2} + s}

\int \frac{1}{x ^{1}} d x

What is the integral of 1/(2x^2-12x+26) ?
The integral of 1/(2x^2-12x+26) is 1/4 arctan((x-3)/2)+C