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

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

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

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

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

Solution steps

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

Complete the square

x^{2} + 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

\frac{\partial}{\partial x} (x^{2} - k x e^{3 y} - 3 y^{2})

f (x) = x^{2} - 5, (2, - 1)

\frac{( 5 x - 4 )}{x - 1}

\frac{d}{d x} (- s t)

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

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