What is the integral of 1/(x^2-8x+17) ?

The integral of 1/(x^2-8x+17) is arctan(x-4)+C

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integral of 1/(x^2-8x+17)

\int \frac{1}{x ^{2} - 8 x + 17} d x

arctan (x - 4) + C

Solution steps

\int \frac{1}{x ^{2} - 8 x + 17} d x

Complete the square

x^{2} - 8 x + 17 : (x - 4)^{2} + 1

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

Apply u-substitution

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

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

Use the common integral:

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

= arctan (u)

Substitute back

u = x - 4

= arctan (x - 4)

Add a constant to the solution

= arctan (x - 4) + C

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What is the integral of 1/(x^2-8x+17) ?
The integral of 1/(x^2-8x+17) is arctan(x-4)+C