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

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

d er i v a t i v e 3 x^{2 x}

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

d er i v a t i v e y = \frac{x}{x + 1}

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

d er i v a t i v e \frac{( 3 x ^{2} )}{( 5 - 3 x ) ^{4}}

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