What is the integral of 1/(x^2-2x+37) ?

The integral of 1/(x^2-2x+37) is 1/6 arctan((x-1)/6)+C

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integral of 1/(x^2-2x+37)

\int \frac{1}{x ^{2} - 2 x + 37} d x

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

Solution steps

\int \frac{1}{x ^{2} - 2 x + 37} d x

Complete the square

x^{2} - 2 x + 37 : (x - 1)^{2} + 36

= \int \frac{1}{( x - 1 ) ^{2} + 36} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 36} d u

= \int \frac{1}{u ^{2} + 36} 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{x - 1}{6})

Add a constant to the solution

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

\frac{d x}{d t} + x^{2} = x

\frac{\partial}{\partial y} (ln (x + 2 y))

f (x) = ln^{2} (x)

\frac{4}{1 + x}

\frac{d}{d x} (sin (2 x^{4}))

What is the integral of 1/(x^2-2x+37) ?
The integral of 1/(x^2-2x+37) is 1/6 arctan((x-1)/6)+C