What is the integral of 1/(x^2-8x+65) ?

The integral of 1/(x^2-8x+65) is 1/7 arctan((x-4)/7)+C

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integral of 1/(x^2-8x+65)

\int \frac{1}{x ^{2} - 8 x + 65} d x

\frac{1}{7} arctan (\frac{x - 4}{7}) + C

Solution steps

\int \frac{1}{x ^{2} - 8 x + 65} d x

Complete the square

x^{2} - 8 x + 65 : (x - 4)^{2} + 49

= \int \frac{1}{( x - 4 ) ^{2} + 49} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 49} d u

= \int \frac{1}{u ^{2} + 49} d u

Apply Integral Substitution

= \int \frac{1}{7 ( 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}{7} \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}{7} arctan (v)

Substitute back

= \frac{1}{7} arctan (\frac{x - 4}{7})

Add a constant to the solution

= \frac{1}{7} arctan (\frac{x - 4}{7}) + C

2 x + 1

\int 5^{- 5 x} d x

(x^{9} - 3 x)^{- 2}

\int (y + 1) d y

\frac{d y}{d x} - \frac{x y}{1 + x} = x

What is the integral of 1/(x^2-8x+65) ?
The integral of 1/(x^2-8x+65) is 1/7 arctan((x-4)/7)+C