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

The integral of 1/(x^2-2x+26) is 1/5 arctan((x-1)/5)+C

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

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

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

Solution steps

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

Complete the square

x^{2} - 2 x + 26 : (x - 1)^{2} + 25

= \int \frac{1}{( x - 1 ) ^{2} + 25} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 25} d u

= \int \frac{1}{u ^{2} + 25} d u

Apply Integral Substitution

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

Substitute back

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

Add a constant to the solution

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

x^{^{''}} - a x = 0

x \to 0 lim (x^{2} cos (\frac{8}{x}))

y = \frac{3 x - 2}{x}

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

\frac{\partial}{\partial t} (4 sin (t))

What is the integral of 1/(x^2-2x+26) ?
The integral of 1/(x^2-2x+26) is 1/5 arctan((x-1)/5)+C