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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

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

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

\frac{v ^{2}}{v ^{2} + 1} = - \frac{1}{v ^{2} + 1} + 1

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int 1 d v = v

= - arctan (v) + v

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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

(\frac{x}{4 x - 3 ln ( x )})^{^{'}}

\frac{5 s + 31}{s ^{2} + 6 s + 9}

\int_{1}^{3} x^{- 2} d x

x \to 2 + lim (\frac{5}{x - 2})

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