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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

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

\int 2 d u = 2 u

= 2 (\frac{u ^{3}}{3} + 2 u)

Substitute back

u = x - 1

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

Simplify

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

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

Add a constant to the solution

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

x \to - 1 lim (8 x^{3})

4 x^{2}, x^{4} - 6 x^{2}, [- 3, 3]

x \to 0 lim (\frac{sin ( π + x )}{x})

y = 12, y = 2 x^{2} - 11 x + 12

f (x) = 2 x^{7} - 3 x^{6} + 3 x^{3} - 4 x^{2} - 7

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