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

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

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

2 (- arctan (x + 1) + x + 1) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

\frac{u ^{2}}{u ^{2} + 1} = - \frac{1}{u ^{2} + 1} + 1

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

\int 1 d u = u

= 2 (- arctan (u) + u)

Substitute back

u = x + 1

= 2 (- arctan (x + 1) + x + 1)

Add a constant to the solution

= 2 (- arctan (x + 1) + x + 1) + C

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

\int \frac{1}{x ( 1 + x ^{7} )} d x

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

\int x 9 + x d x

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