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

The integral of 1/((x+1)*sqrt(x^2+2x)) is arctan(sqrt(x^2+2x))+C

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

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

arctan (x^{2} + 2 x) + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{1}{v ^{2} + 1} d v

= \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= arctan (v)

Substitute back

= arctan ((x + 1)^{2} - 1)

Expand

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

= arctan (x^{2} + 2 x)

Add a constant to the solution

= arctan (x^{2} + 2 x) + C

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

\frac{\partial}{\partial x} (e^{x y z^{5}})

x \to 0 lim (x^{3} - x)

\frac{d}{d x} (\frac{1}{1 - e ^{x}})

(5.7) (2.7)

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