What is the integral of 1/(x^2-6x+34) ?

The integral of 1/(x^2-6x+34) is 1/5 arctan((x-3)/5)+C

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integral of 1/(x^2-6x+34)

\int \frac{1}{x ^{2} - 6 x + 34} d x

\frac{1}{5} arctan (\frac{x - 3}{5}) + C

Solution steps

\int \frac{1}{x ^{2} - 6 x + 34} d x

Complete the square

x^{2} - 6 x + 34 : (x - 3)^{2} + 25

= \int \frac{1}{( x - 3 ) ^{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 - 3}{5})

Add a constant to the solution

= \frac{1}{5} arctan (\frac{x - 3}{5}) + C

y^{^{''}} + y = - sin (3 x)

\frac{d}{d x} (2 sin (π x))

\frac{d}{d x} (e^{\frac{4 x}{3}})

(x^{- 2} + x^{- 3}) \cdot (x^{5} - 2 x^{2})

\frac{d y}{d t} = y (3 - y)

What is the integral of 1/(x^2-6x+34) ?
The integral of 1/(x^2-6x+34) is 1/5 arctan((x-3)/5)+C