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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

\int cos (x) sin (x) x d x

\int (lo g_{10} (x))^{2} d x

\int \frac{ln ( x )}{x} e^{l n^{2} (x)} d x

\int cot^{2} (x) sec (x) d x

\int x^{3} + x^{2} + x d x

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