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

Popular >

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

\frac{d ^{4}}{d x ^{4}} (e^{- x^{2}})

\frac{3}{4 x + 2} + 3 x^{2}

\frac{d}{d x} (e^{- x} x^{2})

x \to - 2 - lim (\frac{x + 5}{x + 4})

\frac{d}{d x} ((x^{2} - 9)^{\frac{2}{3}})

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