# How to you find the general solution of #dy/dx=x/(1+x^2)#?

Multiply both sides by

Integrate:

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To find the general solution of the differential equation ( \frac{dy}{dx} = \frac{x}{1+x^2} ), you can follow these steps:

- Separate the variables by moving all terms involving ( y ) to one side and all terms involving ( x ) to the other side.
- Integrate both sides with respect to their respective variables.
- Solve for ( y ) to find the general solution.

Here's how to do it:

[ \frac{dy}{dx} = \frac{x}{1+x^2} ]

Separate variables:

[ \frac{dy}{1} = \frac{x}{1+x^2} , dx ]

Integrate both sides:

[ \int , dy = \int \frac{x}{1+x^2} , dx ]

[ y = \int \frac{x}{1+x^2} , dx ]

To integrate ( \frac{x}{1+x^2} ), perform a substitution. Let ( u = 1 + x^2 ), then ( du = 2x , dx ), so ( \frac{1}{2} du = x , dx ).

Substitute ( u ) and ( \frac{1}{2} du ) back into the integral:

[ y = \int \frac{1}{u} , \frac{1}{2} du ]

[ y = \frac{1}{2} \int \frac{1}{u} , du ]

[ y = \frac{1}{2} \ln|u| + C ]

Substitute ( u = 1 + x^2 ) back into the equation:

[ y = \frac{1}{2} \ln|1 + x^2| + C ]

So, the general solution to the differential equation is:

[ y = \frac{1}{2} \ln|1 + x^2| + C ]

where ( C ) is the constant of integration.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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