What is the integral of (x+3)/(sqrt(x^2+2x)) ?

The integral of (x+3)/(sqrt(x^2+2x)) is sqrt(x^2+2x)+2ln|sqrt(x^2+2x)+x+1|+C

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integral of (x+3)/(sqrt(x^2+2x))

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

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

Solution steps

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

Complete the square

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

= \int \frac{x + 3}{( x + 1 ) ^{2} - 1} d x

Apply u-substitution

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

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

Expand

\frac{u + 2}{u ^{2} - 1} : \frac{u}{u ^{2} - 1} + \frac{2}{u ^{2} - 1}

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

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

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

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

= u^{2} - 1 + 2 ln u^{2} - 1 + u

Substitute back

u = x + 1

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

Simplify

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

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

Add a constant to the solution

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

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

y^{^{'}} = - 2 y

f (x) = (x^{2} - 1)^{2}, (- 2, 9)

\frac{\partial}{\partial x} (y x^{y - 1})

\frac{6 x ^{2} + 8 x + 8}{x}

What is the integral of (x+3)/(sqrt(x^2+2x)) ?
The integral of (x+3)/(sqrt(x^2+2x)) is sqrt(x^2+2x)+2ln|sqrt(x^2+2x)+x+1|+C