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

The integral of 1/(x^2+2x+65) is 1/8 arctan((x+1)/8)+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+65)

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

\frac{1}{8} arctan (\frac{x + 1}{8}) + C

Solution steps

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

Complete the square

x^{2} + 2 x + 65 : (x + 1)^{2} + 64

= \int \frac{1}{( x + 1 ) ^{2} + 64} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 64} d u

= \int \frac{1}{u ^{2} + 64} d u

Apply Integral Substitution

= \int \frac{1}{8 ( 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}{8} \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}{8} arctan (v)

Substitute back

= \frac{1}{8} arctan (\frac{x + 1}{8})

Add a constant to the solution

= \frac{1}{8} arctan (\frac{x + 1}{8}) + C

\frac{d}{d x} (3 2 x - 5)

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

e^{- t} sin^{2} (t)

x y^{^{'''}} - \frac{2}{x} y^{^{'}} = 0

\frac{d}{d x} (sin (p (x)) i x)

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