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

The solution for 1/x (dy)/(dx)= 1/(x^2+1) is y= 1/2 ln(x^2+1)+c_{1}

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1/x (dy)/(dx)= 1/(x^2+1)

\frac{1}{x} \frac{d y}{d x} = \frac{1}{x ^{2} + 1}

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

Solution steps

\frac{1}{x} \frac{d y}{d x} = \frac{1}{x ^{2} + 1}

Substitute

\frac{d y}{d x}

with

y^{'} (x)

\frac{1}{x} y^{^{'}} (x) = \frac{1}{x ^{2} + 1}

Isolate

y^{'} (x) : y^{'} (x) = \frac{x}{x ^{2} + 1}

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

then

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

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

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

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

f (x) = 5 cos (x)

x \to 0 lim (\frac{x e ^{x}}{sin ( x )})

n = 1 \sum \infty 1 (\frac{5}{6})^{n}

\frac{d}{d x} (\frac{1}{5 x - 2})

k = 1 \sum \infty cos (\frac{1}{k}) - cos (\frac{1}{k + 1})

What is the solution for 1/x (dy)/(dx)= 1/(x^2+1) ?
The solution for 1/x (dy)/(dx)= 1/(x^2+1) is y= 1/2 ln(x^2+1)+c_{1}