What is the integral of 1/(sqrt(x(2+x))) ?

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

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integral of 1/(sqrt(x(2+x)))

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

ln x^{2} + 2 x + x + 1 + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int sec (v) d v

= \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= ln ∣ tan (v) + sec (v) ∣

Substitute back

= ln ∣ tan (arcsec (x + 1)) + sec (arcsec (x + 1)) ∣

Simplify

ln ∣ tan (arcsec (x + 1)) + sec (arcsec (x + 1)) ∣ : ln x^{2} + 2 x + x + 1

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

Add a constant to the solution

= ln x^{2} + 2 x + x + 1 + C

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

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

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

\int x d arctan (x) d x

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

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