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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{v - 1}{v} d v

= 2 \cdot \int \frac{v - 1}{v} d v

Expand

\frac{v - 1}{v} : 1 - \frac{1}{v}

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

\int 1 d v = v

\int \frac{1}{v} d v = ln ∣ v ∣

= 2 (v - ln ∣ v ∣)

Substitute back

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

Add a constant to the solution

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

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

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

\int 6 x - 3 x^{2} d x

\int \frac{3 x ^{2}}{x ^{3} + 4} d x

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

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