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

The integral of 1/((x+1)sqrt(x+2)) is -ln|sqrt(x+2)+1|+ln|sqrt(x+2)-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+1)sqrt(x+2))

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

- ln x + 2 + 1 + ln x + 2 - 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u u + 1} d u

= \int \frac{1}{u u + 1} d u

Apply u-substitution

: \int \frac{2}{v ^{2} - 1} d v

= \int \frac{2}{v ^{2} - 1} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

v^{2} - 1 : - (- v^{2} + 1)

= 2 \cdot \int \frac{1}{- ( - v ^{2} + 1 )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 (- \int \frac{1}{- v ^{2} + 1} d v)

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= 2 (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= 2 (- (\frac{ln x + 1 + 1 + 1}{2} - \frac{ln x + 1 + 1 - 1}{2}))

Simplify

2 (- (\frac{ln x + 1 + 1 + 1}{2} - \frac{ln x + 1 + 1 - 1}{2})) : - ln x + 2 + 1 + ln x + 2 - 1

= - ln x + 2 + 1 + ln x + 2 - 1

Add a constant to the solution

= - ln x + 2 + 1 + ln x + 2 - 1 + C

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

\int 2 x^{2} + 9 d x

\int \frac{1}{36} (x (12 - x)) d x

\int \frac{3}{x ^{2} - 4 x + 1} d x

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

What is the integral of 1/((x+1)sqrt(x+2)) ?
The integral of 1/((x+1)sqrt(x+2)) is -ln|sqrt(x+2)+1|+ln|sqrt(x+2)-1|+C