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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Expand

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

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

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

\int 2 d u = 2 u

\int \frac{1}{u} d u = ln ∣ u ∣

= 2 (\frac{u ^{2}}{2} - 2 u + ln ∣ u ∣)

Substitute back

u = 1 + x + 1

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

Simplify

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

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

Add a constant to the solution

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

\int x^{2} x^{3} + 25 d x

\int \frac{x}{3 - x} d x

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

\int \frac{1}{( x ^{2} + 4 x + 7 ) ^{3}} d x

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

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