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

The integral of 1/((x^2+2x+2)) is arctan(x+1)+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+2x+2))

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

arctan (x + 1) + C

Solution steps

\int \frac{1}{x ^{2} + 2 x + 2} d x

Complete the square

x^{2} + 2 x + 2 : (x + 1)^{2} + 1

= \int \frac{1}{( x + 1 ) ^{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 + 1

= arctan (x + 1)

Add a constant to the solution

= arctan (x + 1) + C

\int \frac{x}{x - 7} d x

\int \frac{49 - y ^{8}}{y ^{5}} d y

\frac{d}{d x} (40 x - 4 x)

\int \frac{1}{( x ^{2} + 16 ) ^{\frac{3}{2}}} d x

f (x) = x^{2} - 3, at x = 3

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