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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Factor

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = x + 1

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

Simplify

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

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

Add a constant to the solution

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

y = ln (\frac{4}{x ^{3}})

\frac{\partial}{\partial x} (x cos (\frac{1}{x ^{2} + y ^{2}}))

\frac{\partial}{\partial x} (e^{(x - y)})

\int \frac{54}{1 - cos ( 6 x )} d x

\frac{d}{d x} (3 x^{2} - 2 x)

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