What is the solution for (dy)/(dx)=((2x+5))/((x^2+2x)),y(1)=1 ?

The solution for (dy)/(dx)=((2x+5))/((x^2+2x)),y(1)=1 is y= 5/2 ln(x)-1/2 ln(x+2)+1+(ln(3))/2

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(dy)/(dx)=((2x+5))/((x^2+2x)),y(1)=1

\frac{d y}{d x} = \frac{( 2 x + 5 )}{( x ^{2} + 2 x )}, y (1) = 1

y = \frac{5}{2} ln (x) - \frac{1}{2} ln (x + 2) + 1 + \frac{ln ( 3 )}{2}

Solution steps

\frac{d y}{d x} = \frac{( 2 x + 5 )}{( x ^{2} + 2 x )}

Substitute

\frac{d y}{d x}

with

y^{'} (x)

y^{^{'}} (x) = \frac{( 2 x + 5 )}{( x ^{2} + 2 x )}

f^{^{'}} (x) = g (x)

then

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

y = \int \frac{2 x + 5}{x ^{2} + 2 x} d x

\int \frac{2 x + 5}{x ^{2} + 2 x} d x = \frac{5}{2} ln (x) - \frac{1}{2} ln (x + 2) + c_{1}

y = \frac{5}{2} ln (x) - \frac{1}{2} ln (x + 2) + c_{1}

Apply initial conditions:

y = \frac{5}{2} ln (x) - \frac{1}{2} ln (x + 2) + 1 + \frac{ln ( 3 )}{2}

y = \frac{5}{2} ln (x) - \frac{1}{2} ln (x + 2) + 1 + \frac{ln ( 3 )}{2}

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

\frac{1}{x}, \frac{1}{x ^{2}}, 1, 2

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

(0, 1), (8, 6)

\frac{3 x - 9}{x}

What is the solution for (dy)/(dx)=((2x+5))/((x^2+2x)),y(1)=1 ?
The solution for (dy)/(dx)=((2x+5))/((x^2+2x)),y(1)=1 is y= 5/2 ln(x)-1/2 ln(x+2)+1+(ln(3))/2