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

The integral of 1/(sqrt(x)(\sqrt{x+1))} is 2ln|sqrt(x+1)+sqrt(x)|+C

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integral of 1/(sqrt(x)(\sqrt{x+1))}

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

2 ln x + 1 + x + 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

Use the common integral:

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

= 2 ln u^{2} + 1 + u

Substitute back

u = x

= 2 ln (x)^{2} + 1 + x

(x)^{2} = x

= 2 ln x + 1 + x

Add a constant to the solution

= 2 ln x + 1 + x + C

\int \frac{1}{( x ^{2} + 81 ) ^{\frac{3}{2}}} d x

x \to 8 lim (\frac{( x ^{2} - 5 x - 24 )}{x - 8})

\int (x + a)^{2} d x

7 + cot (x) - 2 csc (x)

\frac{d y}{d x} = - 3 - 2 x - y

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