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

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

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

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

ln \frac{x + 1}{x} + 1 - ln \frac{x + 1}{x} - 1 - 2 \frac{x + 1}{x} + C

Solution steps

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

\frac{1}{x} = \frac{1 - x}{x} + 1

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{2 v ^{2}}{v ^{2} - 1} d v

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

Take the constant out:

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

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

Factor

v^{2} - 1 : - (- v^{2} + 1)

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

Take the constant out:

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

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

\frac{v ^{2}}{- v ^{2} + 1} = \frac{1}{- v ^{2} + 1} - 1

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

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

\int 1 d v = v

= - 2 (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2} - v))

Substitute back

= - 2 - \frac{ln 1 + \frac{1 - x}{x} + 1 + 1}{2} - \frac{ln 1 + \frac{1 - x}{x} + 1 - 1}{2} - 1 + \frac{1 - x}{x} + 1

Simplify

- 2 - \frac{ln 1 + \frac{1 - x}{x} + 1 + 1}{2} - \frac{ln 1 + \frac{1 - x}{x} + 1 - 1}{2} - 1 + \frac{1 - x}{x} + 1 : ln \frac{x + 1}{x} + 1 - ln \frac{x + 1}{x} - 1 - 2 \frac{x + 1}{x}

= ln \frac{x + 1}{x} + 1 - ln \frac{x + 1}{x} - 1 - 2 \frac{x + 1}{x}

Add a constant to the solution

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

\int \frac{5 x - 4}{x ^{3} + 4 x} d x

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

\int (5 - 4 x) e^{6 - 5 x} d x

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