What is the integral of 1/(x^2+10x+29) ?

The integral of 1/(x^2+10x+29) is 1/2 arctan((x+5)/2)+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+10x+29)

\int \frac{1}{x ^{2} + 10 x + 29} d x

\frac{1}{2} arctan (\frac{x + 5}{2}) + C

Solution steps

\int \frac{1}{x ^{2} + 10 x + 29} d x

Complete the square

x^{2} + 10 x + 29 : (x + 5)^{2} + 4

= \int \frac{1}{( x + 5 ) ^{2} + 4} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 4} d u

= \int \frac{1}{u ^{2} + 4} d u

Apply Integral Substitution

= \int \frac{1}{2 ( 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}{2} \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}{2} arctan (v)

Substitute back

= \frac{1}{2} arctan (\frac{x + 5}{2})

Add a constant to the solution

= \frac{1}{2} arctan (\frac{x + 5}{2}) + C

f (x) = (arcsec (x))^{13}

\int 3 x sin^{2} (x) d x

\int_{0}^{1} - \frac{1}{x - 5} + \frac{2}{x - 2} d x

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

\int (2 x + 6)^{5} d x

What is the integral of 1/(x^2+10x+29) ?
The integral of 1/(x^2+10x+29) is 1/2 arctan((x+5)/2)+C