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

The integral of (sqrt(1+x)+1)/(sqrt(1+x)-1) is x+4ln|sqrt(x+1)-1|+4sqrt(x+1)-4+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (sqrt(1+x)+1)/(sqrt(1+x)-1)

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

x + 4 ln x + 1 - 1 + 4 x + 1 - 4 + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{( v + 1 ) ( v + 2 )}{v} d v

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

Expand

\frac{( v + 1 ) ( v + 2 )}{v} : v + 3 + \frac{2}{v}

= 2 \cdot \int v + 3 + \frac{2}{v} d v

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 v d v + \int 3 d v + \int \frac{2}{v} d v)

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

\int 3 d v = 3 v

\int \frac{2}{v} d v = 2 ln ∣ v ∣

= 2 (\frac{v ^{2}}{2} + 3 v + 2 ln ∣ v ∣)

Substitute back

= 2 (\frac{( 1 + x - 1 ) ^{2}}{2} + 3 (1 + x - 1) + 2 ln 1 + x - 1)

Simplify

2 (\frac{( 1 + x - 1 ) ^{2}}{2} + 3 (1 + x - 1) + 2 ln 1 + x - 1) : x + 4 ln x + 1 - 1 + 4 x + 1 - 4

= x + 4 ln x + 1 - 1 + 4 x + 1 - 4

Add a constant to the solution

= x + 4 ln x + 1 - 1 + 4 x + 1 - 4 + C

\int \frac{tan ( x ) + sin ( x )}{sec ( x )} d x

\int 6 x e^{5 x} d x

\int x^{2} (5 x^{3} - 1)^{4} d x

\int \frac{x + 1}{x ^{2} - 16} d x

n = 0 \sum \infty 4 n^{2} + 4 n + 1

What is the integral of (sqrt(1+x)+1)/(sqrt(1+x)-1) ?
The integral of (sqrt(1+x)+1)/(sqrt(1+x)-1) is x+4ln|sqrt(x+1)-1|+4sqrt(x+1)-4+C