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 ) 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 \to 0 + lim (x^{5 s i n (x)})

\frac{\partial}{\partial z} (4 x ln (y) + y ln (z) - x)

(x - 4 y - 3) d x - (x - 6 y - 5) d y = 0

n \to \infty lim (cos (n))

x \to 0 lim (\frac{( sin ( 9 x ) )}{x})

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