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

The integral of (x+1)/(sqrt(x)-1) is 2/3 x^{3/2}+x+4sqrt(x)+4ln|sqrt(x)-1|+C

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

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

\frac{2}{3} x^{\frac{3}{2}} + x + 4 x + 4 ln x - 1 + C

Solution steps

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

Apply u-substitution

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

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

Long division

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

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

\int u^{2} d u = \frac{u ^{3}}{3}

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

\int 2 d u = 2 u

\int \frac{2}{u - 1} d u = 2 ln ∣ u - 1 ∣

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

Substitute back

u = x

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

Simplify

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

= \frac{2}{3} x^{\frac{3}{2}} + x + 4 x + 4 ln x - 1

Add a constant to the solution

= \frac{2}{3} x^{\frac{3}{2}} + x + 4 x + 4 ln x - 1 + C

\int (4 x - \frac{2}{x} + 5 sin (x)) d x

x \to \infty lim ((4)^{x})

f^{^{'}} (x) = sec (2 x) tan (2 x) - \frac{4}{x ^{2}}

40 ln (2 x + 1)

n = 0 \sum \infty x^{n} \frac{1}{n}

What is the integral of (x+1)/(sqrt(x)-1) ?
The integral of (x+1)/(sqrt(x)-1) is 2/3 x^{3/2}+x+4sqrt(x)+4ln|sqrt(x)-1|+C